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\begin{document}
\title{$p$-adic abelian integrals}
\author{Pierre Colmez}
\date{Recorded by Shenxing Zhang and revised by Ruide Fu}
\begin{abstract}
The study of complex abelian integrals, i.e., integrals of algebraic functions of one complex variable, was a major incentive to develop complex algebraic geometry (some 150 years ago).  After briefly explaining the complex theory, I will study its analog in the p-adic world: this provides a concrete introduction to p-adic Hodge theory, a theory that was originated by Tate some 50 years ago and was turned into one of most powerful tools of number theory.

This is the note of the lectures in BICMR, Beijing from 2016/09/14 to 2016/10/26.
\end{abstract}
\maketitle
\tableofcontents
\vspace{0.4cm}

\section{Complex abelian integral on elliptic curves}

\subsection{Building blocks of functions on $\BC$ associate to a lattice}
Let $E/\BC$ be an elliptic curve given by a Weierstrass equation
  \begin{equation}
    y^2=4x^3-g_2x-g_3,
  \end{equation}
$\Lambda$ be the image of $\RH_1(E(\BC),\BZ)$ in $\BC$ by
  \[ u\mapsto \int_u \frac{\diff x}{y}.\]
Then we have an isomorphism of Riemann surfaces, through which we can define an addition on $E$, induced by addition on $\BC$:
  \begin{equation}
    \Function{\alpha: E}{\BC/\Lambda,}{P}{\int_O^P\frac{dx}{y}.}
  \end{equation}
The inverse is given by
  \begin{equation}
    \Phi_\Lambda: z\longmapsto (\wp,\wp'),
  \end{equation}
where the Weierstrass $\sigma$, $\zeta$ and $\wp$ functions are defined as
  \begin{eqnarray}
    \sigma(z,\Lambda)&=&z\prod_{w\in \Lambda-\{0\}} (1-\frac{z}{w})e^{\frac{z}{w}+\frac{z^2}{2w^2}},\\
    \zeta(z,\Lambda)&=&\frac{\diff}{\diff z}\log\sigma(z,\Lambda)=\frac{1}{z}+\sum_{w\in\Lambda-\{0\}}(\frac{1}{z-w}+\frac{1}{w}+\frac{z}{w^2}),\\
    \wp(z,\Lambda)&=&-\frac{\diff}{\diff z}\zeta(z,\Lambda)=\frac{1}{z^2}+\sum_{w\in\Lambda-\{0\}}(\frac{1}{(z-w)^2}-\frac{1}{w^2}).
  \end{eqnarray}
  
\begin{prop}
Fix a lattice $\Lambda$, and let $w \in\Lambda$, we then have the formulae
	\[\sigma(z+w)=\sigma(z)\exp(\eta(w)z+\theta(w)),\]
where $\eta$ and $\theta$ are constants depending on $w$.
\end{prop}
\begin{proof}
This argument is a consequence of
	\[\begin{split}
		\dlog\frac{\sigma(z+w)}{\sigma(z)}&=\zeta(z+w)-\zeta(z)\\
		&=\int_z^{z+w}-\wp(\xi)\diff\xi,
	\end{split}\]
and that the last integral does not depend on $z$ if $w$ is in $\Lambda$, denoted by $\eta(w)$.
\end{proof}
  
\begin{prop}
The field of rational functions on $\BC/\Lambda$ is generated by $\wp$ and $\wp'$.
\end{prop}

\subsection{Abel theory}
Let $D\in\Div(\BC/\Lambda)=\BZ[\BC/\Lambda]$ be a divisor on $\BC/\Lambda$, then
  \[D=\sum_{w\in \BC/\Lambda} n_w [w],\quad n_w\in\BZ,\]
$n_w=0$ for almost all $w$. Define
  \[\begin{split}
    \deg D&=\sum_w n_w,\\
    \Tr D&=\sum n_w w\in \BC/\Lambda.
  \end{split}\]
Denote by $\Div^0(\BC/\Lambda)$ the subgroup of $\Div(\BC/\Lambda)$ consisting of all degree zero divisors. For any rational function $f\in \BC(\BC/\Lambda)^\times$, define
  \[\div(f)=\sum v_w(f) w,\]
where $v_w$ is the order of $f$ at $w$.
\begin{thm}[Abel]
$\deg D=0$ and $\tr D=0$ if and only if $D=\div(f)$ for some $f\in\BC(\BC/\Lambda)^\times$.
\end{thm}

\begin{prop}
Let $D=\sum n_i[z_i]$ be a divisor on $\BC$ such that $\sum n_i=0$ and $\sum n_iz_i=0$, then
  \[ \prod \sigma(z-z_i,\Lambda)^{n_i}\]
is a rational function on $\BC/\Lambda$ with divisor $\bar D=\sum n_i(\bar{z_i})$.
\end{prop}
\begin{coro}
We hence have an isomorphism $E_\Lambda\simeq\frac{\Div(\BC/\Lambda)}{\Div(f)}$.
\end{coro}
\begin{thm}
(i) For any $f\in\BC(E)$, $\Phi_\Lambda^*(f)=f\circ\Phi_\Lambda$ can be written uniquely as
  \[\lambda_0+\sum_{i=1}^n\sum_{k=1}^{k_i}\frac{\lambda_{i,k}}{k!}\zeta^{(k-1)}(z-a_i,\Lambda),\]
where $\lambda_0,...\lambda_{i,k}\in\BC$, $a_i\in \BC\mod\Lambda,$ $\sum\lambda_{i,1}=0.$ Conversely, such expression is $\Phi^*_\Lambda f$  for some $f\in \BC(E)$ if $\sum \lambda_{i,1}=0$.

(ii) The integration of $f\in\BC(E)$ is given by 
  \[\int f\circ \phi_\Lambda=\lambda_0 z+\sum_{i=1}^n \lambda_{i,1}\log\sigma(z-a_i)+\sum_{i=1}^n \sum_{k=2}^{k_i}\frac{\lambda_{i,k}}{k!}\zeta^{(k-2)}(z-a_i),\]
in the complex plane, and is a rational function on $E_\Lambda$ if and only if $\lambda_0=0$, $\lambda_{i,1}=0$ for all $i$, and $\sum \lambda_{i,2}=0$.
\end{thm}

\subsection{Rational differential forms on $E$}
For $f\in \BC(E)$, let $\omega=f\frac{\diff x}{y}\in\Omega^1_{\BC(E)}$ be a rational differential on $E$. Then
  \[ \phi_\Lambda^*\omega=(f\circ\phi_\Lambda)\diff z. \]
\begin{defn}
We say $\omega$ is of the
\begin{itemize}
\item \emph{first kind} if it is holomorphic ($\iff f\circ \phi_\Lambda$ is constant);
\item \emph{second kind} if it has no residue ($\iff \lambda_{i,1}=0$ for all $i$);
\item \emph{third kind} if it only has simple poles and residues in $\BZ$ ($\iff k_i=1$ and $\lambda_{i,1}\in\BZ$ for all $i$).
\end{itemize}
Denote by $\RH^0(E,\Omega^1),\DSK(E),\DTK(E)$ the three kind of differential forms respectively. Then
  \[\RH^0(E,\Omega^1)=\BC\frac{\diff x}{y}\]
and
  \[\begin{split}
    \DSK(E)&\supseteq \{\diff f: f\in \BC(E)\},\\
    \DTK(E)&\supseteq \{\frac{\diff f}{f}: f\in \BC(E)^\times\},
  \end{split}\]
the right hand sides are called exact forms.

\end{defn}

Let $u$ be a path on $E(\BC)$. For $\omega\in\DSK(E)$, $\int_u\omega$ depends only on the image of $u$ in $\RH_1(E(\BC),\BZ)$. For $\omega\in\DTK(E)$, $\int_u\omega\mod 2\pi i \BZ$ depends only on the image of $u$ in $\RH_1(E(\BC),\BZ)$.

For $\omega\in\DTK(E)$,
  \[\phi_\Lambda^*\omega=(\lambda_0+\sum_{i=1}^n \lambda_{i,1}\zeta(z-a_i,\Lambda))\diff z.\]
Denote
  \begin{equation}
    \div(\omega)=\sum_{i=1}^n \lambda_{i,1}(\phi_\Lambda(a_i))\in \Div^0(E).
  \end{equation}
Then we have an exact sequence
  \[0\ra \RH^0(E,\Omega^1)\ra \DTK(E)\ra \Div^0(E)\ra 0.\]
Notice that for $f\in \BC(E)^\times$, $\div(\frac{\diff f}{f})=\div(f)$. By Abel's theorem,
  \[\begin{split}
    \frac{\Div^0(E)}{\{\div(f)\}}&\simto E(\BC)\\
    \sum n_iP_i&\mapsto \oplus n_i P_i.
  \end{split}\]
Hence we have a commutative diagram with exact rows and columns:
  \[\xymatrix{
    & & \{\frac{\diff f}{f}\} \ar[r]^-\sim \ar[d] &\{\div(f)\}\ar[d] &\\
    0\ar[r]&\RH^0(E,\Omega^1)\ar@{=}[d]\ar[r]&\DTK(E)\ar[r]\ar[d]&\Div^0(E)\ar[r]\ar[d]&0\\
    0\ar[r]&\RH^0(E,\Omega^1)\ar[r]&\DTK(E)/\{\frac{\diff f}{f}\}\ar[r]& E(\BC)\ar[d]\ar[r]&0\\
    &&&0
  }\]
The group $\RH^0(E,\Omega^1)$ on the last line is an algebraic group denoted $\Ga$. It is simply $\BC$ in our case. The elliptic curve $E(\BC)$ on the last line is also an algebraic group.

It turns out that $\DTK(E)/{\frac{\diff f}{f}}$ can be made an algebraic group as well, which is called the universal extension of $E$.

\begin{defn}
For any $\omega_1$, $\omega_2\in\Lambda$, the \emph{intersection number} $\omega_1\#\omega_2$ is the discriminant of $(\omega_1,\omega_2)$ under an orientable basis of $\Lambda$. That is to say, for a basis $\{w_1,w_2\}$ of $\Lambda$ with $\Im(w_2/w_1)>0$,
  \[u\# v=\det(\int_u \frac{\diff x}{y},\int_v \frac{\diff x}{y}).\]
\end{defn}
\begin{thm}\begin{enumerate}
\item $\frac{\diff x}{y},\frac{x\diff x}{y}\in \DSK(E)$.
\item $\omega\in\DSK(E)$ is exact if and only if $\int_u \omega=0$ for any $u$.
\item We have the Legendre relation. For $u,v\in \RH_1(E(\BC),\BZ)$,
  \[\int_u\frac{\diff x}{y}\int_v\frac{x\diff x}{y}-\int_u\frac{x\diff x}{y}\int_v\frac{\diff x}{y}=2\pi iu\#v.\]
\item $H_{\dR}^1(E):=\DSK(E)/{\{\diff f\}}$ is of dimension $2$, which is generated by $\{\frac{\diff x}{y},\frac{x\diff x}{y}\}$.
\end{enumerate}\end{thm}
\begin{remark}
Assume $E$ is defined over $\ov\BQ$. If $E$ has complex multiplication (CM), then
  \[\ov \BQ(\int_u\frac{\diff x}{y},\int_u\frac{x\diff x}{y}: u\in\RH_1(E(\BC),\BZ))\]
has transcendental degree $2$. It's conjecturally that if $E$ doesn't have CM, the transcendental degree should be $4$. That's Grothendieck's ``Hodge conjecture is false for trivial residues''.
\end{remark}
\begin{proof}
(1) That's because \[\phi_\Lambda^*\frac{\diff x}{y}=\diff z,\quad \phi^*_\Lambda\frac{x\diff x}{y}=\wp(z)\diff z=\zeta'(z)\diff z.\]

(2) Suppose $\phi_\Lambda^*\omega=\diff F$ on $\BC$, then $F(w)=\int_a^w \phi_\Lambda^* \omega$ does not depend on the choice of path and then $\int_u\omega=0$

If $\int_u\omega=0$ for any $u$, then $F(w)=\int_a^w \phi_\Lambda^* \omega$ does not depend on the choice of path. Moreover, $F(z+w)=F(z)$ for any $w\in\Lambda$. Hence $F$ is an elliptic function and then $F=\phi_\lambda^* f$ for some $f\in\BC(E)$. Therefore $\omega=\diff f$.

(3) By bilinearity, we may assume $\{u,v\}$  is a basis of $\RH_1(E(\BC),\BZ)$ and $\int_u\frac{\diff x}{y}, \int_v\frac{\diff x}{y}$ is an oriented basis. The integration of $\zeta(z)$ on the polygon with counterclockwise  vertices $a,a+w_1,a+w_1+w_2,a+w_2,a$ is
  \[\int \zeta(z)\diff z=2\pi i.\]
Meanwhile, it is
  \[\begin{split}&\int_{a}^{a+w_1}(\zeta(z)-\zeta(z+w_2))\diff z-\int_{a}^{a+w_2}(\zeta(z)-\zeta(z+w_1))\diff z\\
  =&\int_{a}^{a+w_1}\int_z^{z+w_2}\wp(\tau)\diff \tau\diff z-\int_{a}^{a+w_2}\int_z^{z+w_1}\wp(\tau)\diff \tau\diff z\\
  =&\int_u\frac{\diff x}{y}\int_v\frac{x\diff x}{y}-\int_u\frac{x\diff x}{y}\int_v\frac{\diff x}{y}.
  \end{split}\]

(4) This follows from (2) and (3).
\end{proof}


\begin{thm}
The pairing
  \[ \Function{(\RH_1(E(\BC),\BZ)\otimes\BC)\times \RH_{\dR}^1(E)}{\BC}{(u,\omega)}{\int_u\omega}\]
is perfect.
\end{thm}


\subsection{Algebraic universal extension}

\begin{prop}
For any $w\in\Lambda$,
  \[\frac{\zeta(z+w,\Lambda)}{\zeta(z,\Lambda)}=\pm e^{\eta(w,\Lambda)(z+\frac{w}{2})},\]
where $\eta(w,\Lambda)=\zeta(z+w,\Lambda)-\zeta(z,\Lambda)$ and the sign depends on whether $\frac{w}{2}$ is in $\Lambda$.
\end{prop}


Let $E/\BC$ be an elliptic curve.
Denote by $m,\pr_1,\pr_2:E\times E\ra E$ the morphism $m(x,y)=x+y, \pr_1(x,y)=x,\pr_2(x,y)=y$. For any $\omega\in\Omega^1_{E/\BC}$, denote by
  \[\delta \omega=m^*\omega-\pr_1^*\omega-\pr_2^* \omega.\]
For any $f\in \BC(E\times E)$, denote by
  \[\delta F(x,y)=F(x\oplus y)-F(x)-F(y).\]
\begin{thm}[Theorem of the square]
\begin{enumerate}
\item If $\omega\in\DSK(E)$, there exists a unique $F\in \BC(E\times E)$ up to constant such that $\delta \omega=\diff F$.
\item If $\omega\in\DTK(E)$, there exists a unique $F\in \BC(E\times E)^\times$ up to constant such that $\delta \omega=\frac{\diff F}{F}$.
\end{enumerate}
\end{thm}
\begin{proof}
(1) Let $\diff F_\omega$ be the pullback of $\phi_\Lambda^*\omega$ on $\BC$, then $\diff \delta F_\omega$ is a pullback of $(\phi_\Lambda\times \phi_\Lambda)^* \delta\omega$ on $\BC\times \BC$. We want to prove that
  \[\delta F_\omega=F_\omega(z_1+z_2)-F_\omega(z_1)-F_\omega(z_2)\]
is periodic of  period $\Lambda\times\Lambda$. Write
  \[\phi_\Lambda^* \omega=(\lambda_0+\sum_{i=1}^n \sum_{k=1}^{k_i}\frac{\lambda_{i,k}}{k!}\zeta^{(k)}(z-a_i,\Lambda))\diff z,\]
then
  \[F_\omega=\lambda_0 z+\sum_{i=1}^n \sum_{k=1}^{k_i}\frac{\lambda_{i,k}}{k!}\zeta^{(k-1)}(z-a_i,\Lambda).\]
If $k\ge 2$, $\zeta^{(k-1)}$ is already periodic. Since $\zeta(z+w)-\zeta(z)=\eta(w)$ if $w\in\Lambda$,
  \[G_i(z_1,z_2)=\zeta(z_1+z_2-a_i)-\zeta(z_1-a_i)-\zeta(z_2-a_i)\]
is periodic of period $\Lambda\times\Lambda$.

(2) Write
  \[\phi_\Lambda^* \omega=(\lambda_0+\sum_{i=1}^n \lambda_i \zeta(z-a_i,\Lambda))\diff z,\quad \lambda_i\in\BZ,\sum \lambda_i=0.\]
Then we need to show that if $f(z)=\frac{\sigma(z-a)}{\sigma(z-b)}$, then $\frac{f(z_1+z_2)}{f(z_1)f(z_2)}$ is periodic of period $\Lambda\times\Lambda$. But this follows from $\sigma(z+w)=e^{a(w)z+b(w)}\sigma(z)$.
\end{proof}

\begin{thm}
There is an algebraic group $\wt E$ (called the universal extension of $E$) with
\begin{enumerate}
\item exact sequence of algebraic groups
  \[0\ra \Ga\ra\wt E\ra E\ra 0.\]
\item $\wt E(\BC)=\DTK(E)/\{\frac{\diff f}{f}\}$ as a group.
\item The following diagram commutes and the rows are exact:
  \[\xymatrix{
     & 0\ar[r]&\BC \ar[d]_\wr\ar[r]^-{i_2}&\frac{\BC\times\BC}{\{(w,\eta(w)):w\in\Lambda\}}\ar[d]_\wr^\phi\ar[r]^-{\pr_1}&\BC/\Lambda\ar[d]_\wr\ar[r]&0\\
     & 0\ar[r]&\BC \frac{\diff x}{y}\ar[r]&\wt E(\BC)\ar[r]^{\pi}&E(\BC)\ar[r]&0
  }\]
where
		\[
			\phi(z_1, z_2) \mapsto (\zeta(z - z_1) - \zeta(z) + z_2) \diff z
		\]
is an isomorphism of groups. Moreover, the first row is exact as algebraic groups.
\item $\pi^*(x\frac{\diff y}{y})+\diff z_2=\diff F$ for some rational function $F$ on $\wt E$.  Thus $\text{H}^1_{\text{dR}}(E)$ can be identified to the invariant differentials 		on $\widetilde E$.
\end{enumerate}\end{thm}
\begin{proof}

			We first define $\widetilde E \simeq \mathbb{C} \times \mathbb{C} / {(w, \eta(w))}$
			as an algebraic variety.
			
			For a point $a$ on $E(\mathbb{C})$ and $\tilde a$ a lifting of it,
			we define a map
			\begin{align}
				(\mathbb{C} - \{\tilde a + \Lambda\}) \times \mathbb{C}
				&\longrightarrow \mathbb{C} \times \mathbb{C} \\
				(x, \lambda) &\longrightarrow (x, \zeta(x - \tilde a) - \zeta(-\tilde a) + \lambda).
			\end{align}
			Note the image of $(x, \lambda)$ and $(x + w, \lambda)$ differ by
			$(w, \eta(w))$ provided $w \in \Lambda$, so this map induces a map
			$s_a: U_a \times \mathbb{C} \to \mathbb{C} \times \mathbb{C} / (w, \eta(w))$,
			where $U_a$ stands for $E(\mathbb{C}) - a$.
			
			For another point $b$ on $E(\mathbb{C})$, we similarly have a map
			$s_b: U_b \times \mathbb{C} \to$ \mbox{$\mathbb{C} \times \mathbb{C}$} $/ (w, \eta(w))$.
			Let $f_{a, b}(x) =
				\zeta(\tilde x - \tilde a) - \zeta(-\tilde a) - \zeta(\tilde x - \tilde b) + \zeta(-\tilde b)$,
			then the map $(x, \lambda) \mapsto (x, \lambda + f_{a, b}(x))$
			induces an algebraic function $\phi_{a, b}$ on $(U_a \cap U_b) \times \mathbb{C}$,
			with the property that $s_a = s_b \circ \phi_{a, b}$.
			
			Now we show that $\mathbb{C} \times \mathbb{C} / (w, \eta(w))$ is an algebraic group.
			In fact, the addition law on $\mathbb{C} \times \mathbb{C} / (w, \eta(w))$ induces an
			addition on $U_a \times \mathbb{C}$, whose formulae is given by
			\[
				(x, \lambda) + (x', \lambda') = (x \oplus x', \lambda + \lambda' + G(x, x')),
			\]
			where $G(x, x')$ is an algebraic function induced by
			\[
				\zeta(x - \tilde a) + \zeta(x' - \tilde a) - \zeta(-\tilde a) - \zeta(x + x' - \tilde a).
			\]
			
			The isomorphism $\widetilde E \simeq \text{DTK}(E)$ is defined locally by
			$\psi_a: U_a \times \mathbb{C} \to \text{DTK}(E)$,
			\[
				(x ,\lambda) \mapsto (-\zeta(z + \tilde x - \tilde a) + \zeta(z - \tilde a)
					+ \zeta(\tilde x - \tilde a) - \zeta(-\tilde a) + \lambda) \text{d}z.
			\]
			Note the result of the mapping is independent of the choice of $\tilde x$ and $\tilde a$.
			Furthermore, this locally defined map is in fact global since we have
			\[
				\psi_a(x, \lambda) - \psi_b(x, \lambda + f_{a, b}(x)) =
					\dlog \frac{\sigma(z + \tilde x - \tilde b) \sigma(z - \tilde a)}
					{\sigma(z + \tilde x - \tilde a) \sigma(z - \tilde b)},
			\]
			in which the right hand side is the logarithm derivative of a function on $E(\mathbb{C})$.
\end{proof}
%\begin{remark}$\wt E$ is called the universal extension of $E$, there is an algebraic definition of Eisenstein series of weight one using this universal extension, see Appendix~\ref{app:eisen}.\end{remark}


\subsection{Weil pairing}
Let $E$ be an elliptic curve over a filed $K$ of characteristic $0$. Let
		$G_K = \text{Gal}(\bar K / K)$. Then for any integer $m \geq 1$,
		$E[m] \simeq (\mathbb{Z} / m \mathbb{Z})^2$ and this gives
		\[
			\rho_{E, m}: G_K \to \text{GL}_2(\mathbb{Z} / m \mathbb{Z}).
		\]
		The first statement follows from that $E$ is defined over $\mathbb{Q}(g_2, g_3)$,
		which can be identified with a subfield of $\mathbb{C}$. This is an example of
		Lefschetz principle, which proposes that an algebraic statement over algebraic
		closed filed of characteristic zero can be checked by just looking at $\mathbb{C}$.
		
		The representations $\rho_{E, m}$ are very interesting. For $p \geq 5$ and
		$E: y^2 = x(x - a^p)(x + b^p)$, then $\rho_{E, m}$ has so nice property that
		\[
			a^p + b^p = c^p, \quad (a, b, c) = 1,
		\]
		cannot have integral solution.

\begin{thm}
\begin{enumerate}
\item For any $P\in E[m]$, there is a unique $f\in\bar K(E)^\times$ up to $\bar K^\times$ such that $\div(f)=m([P]-[O])$.
\item For $P,Q\in E[m]$,
  \[e_m(P,Q)=\frac{f_Q(x)}{f_Q(x\ominus P)}\frac{f_P(x\ominus Q)}{f_P(x)}\in\mu_m\]
is constant. 
\item Moreover, $(P,Q)\mapsto e_m(P,Q)$ gives a bilinear, alternating, non-degenerated pairing on $E[m]\times E[m]$.
\item If $K=\BC$, and $\phi_\Lambda:\BC/\Lambda\simto E(\BC)$, then
  \[e_m(P,Q)=e^{\frac{2\pi i}{m} ma\#mb},\]
where $a,b$ is an inverse image of $P$ and $Q$ in $\BC$.
\end{enumerate}
\end{thm}
\begin{proof}
Assume $K=\BC$ and let $\phi_\Lambda:\BC/\Lambda\simto E(\BC)$. The uniqueness follows from the fact that a regular function  on $E$ without poles and zeroes must be constant.

By Abel's theorem,
  \[f_P(z)=\sigma(z-a)^m\sigma(z)^{1-m}\sigma(z-ma)^{-1}\]
is a rational function on $E(\BC)$ with divisor $m([P]-[O])$. Then
  \[\begin{split}&e_m(P,Q)=\frac{\sigma(z-a-mb)}{\sigma(z-a)}\cdot \frac{\sigma(z-b)}{\sigma(z-b-ma)}\cdot\frac{\sigma(z-ma)}{\sigma(z-mb)}\\
    =&\exp(\frac{ma \eta(mb)-mb\eta(ma)}{m})=\exp({\frac{2\pi i}{m}(ma\# mb)}).\qedhere
  \end{split}\]
\end{proof}

\section{Complex abelian integral on algebraic curves}
\subsection{Algebraic curve over $\BC$}
An curve $X$ over $\BC$ is called proper if $X(\BC)$ is compact; projective if it is defined by a homogeneous polynomial; smooth if  locally holomorphic to an open disk. Thus a smooth and proper algebraic curve $X$
		over $\mathbb{C}$ gives a compact Riemann surface $X(\mathbb{C})$, and vice versa
		(hard!). Let $g$ be its genus. Then topologically it's a  $4g$-gon with edges identified. %For general definition and properties of intersection numbers on a orientable topological manifold please see appendix.
		
Fix a point $P_0$ on $X(\mathbb{C})$, the corresponding fundamental group is 
		\[
			\pi_1(X(\mathbb{C}), P_0) = <a_i, b_i, i = 1, \dots, g
			| \prod_{i = 1}^g a_i b_i a_i^{-1} b_i^{-1} = 1>,
		\]
		and the fist homology group is the abelianization of it.

\begin{figure}[h]
  \centering
  % Requires \usepackage{graphicx}
  \includegraphics[width=150pt]{4ggon.png}\\
  \caption{$4g$-gon}
\end{figure}

The intersection pairing
  \[\Function{H_1(X(C),\BZ)\times H_1(X(\BC),\BZ)}{\BZ}{(a,b)}{a\#b}\]
is a bilinear alternating paring. There exist a canonical basis $\{a_1,...,a_g,b_1,...,b_g\}$ of $\RH_1(X(C),\BZ)$ such that
  \[a_i\#b_j=\delta_{ij}=-b_j\#a_i,\quad a_i\#a_j=0=b_i\#b_j.\]
That is to say, under the basis $\{a_1,\dots,a_g,b_1,\dots,b_g\}$, the matrix of intersection numbers is
  \[\begin{pmatrix}
    O& I_g\\-I_g&O
  \end{pmatrix}.\]
Topologically, $a_i$ and $b_i$ are the sides of a $4g$-gon.
This also holds for compact orientable topological manifold.

%$(Riemann's\ period\ relation)$ For any $\xi$, $\eta\in H_{dR}^1(X(C))=\frac{\{closed\ 1-forms\}}{\{exact\ 1-forms\}},$
%$$\int_{X(C)}\xi\wedge\eta=\sum_{i=1}^g (\int_{a_i}\xi\int_{b_i}\eta-\int_{a_i}\eta\int_{bi}\xi).$$

%The above fact holds for compact orientable smooth surface.

\begin{thm}\begin{enumerate}
\item $\dim_{\BC}\RH^0(X(\BC),\Omega_X^1)=g.$
\item There exists a (unique) basis $(\omega_1,...,\omega_g)$ of $\RH^0(X(\BC),\Omega_X^1)$ such that $\int_{a_i}\omega_j=\delta_{ij}$.
\item The matrix $B=(z_{ij})_{1\leq i,j\leq g}=(\int_{b_i}\omega_j)$ is symmetric and $\Im B$ is positive definite.
\end{enumerate}\end{thm}
Let $\Lambda=\BZ^g\oplus B\BZ^g\subset\BC^g$ be the image of $\RH_1(X(\BC),\BZ)$ by
  \[u\mapsto \int_u \ul\omega=(\int_u\omega_1,...,\int_u\omega_g)\]
and $J(\BC)=\BC^g/\Lambda$ be a complex torus.
Fix a point $P_0\in X(\BC)$, the map
  \begin{equation}\iota_{P_0}(P)=\int_{P_0}^P\ul \omega\mod\Lambda\end{equation}
fits in the following commuting  diagram
  \[\xymatrix{
    \pi_1(X(\BC),P_0)\ar[d]\ar[r]^-{\iota_{P_0}} &\pi_1(J(\BC),0)\ar[d]^-{\simeq}\\
    \RH_1(X(\BC),\BZ)\ar[r]^-\simeq&\Lambda
  }\]

\begin{thm}[Riemann]\begin{enumerate}
\item $J$ has a unique structure of algebraic projective variety over $\BC$ of dimension $g$ and $J(\BC)=\BC^g/\Lambda$ endows $J(\BC)$ with a group law, which gives a algebraic group structure of $J$.
\item $\iota_{P_0}$ gives an embedding of algebraic varieties.
\item The induced morphism $\iota_{P_0}^*: \RH^0(J,\Omega^1)\to \RH^0(X,\Omega^1)$ is an isomorphism and $\iota_{P_0}^*dz_i=\omega_i$.
\end{enumerate}\end{thm}
\begin{remark}
(1) $J$ is called the Jacobian of $X$. If $X$ is defined over a number field $K$, then so is $J$.

(2) If $g\le 1$, then $\iota_{P_0}$ is an isomorphism. But for $g\ge 2$, $X$ is very small in $J$.

(3) $J$ is very useful to study $X$. The Mordell-Weil theorem says that $J(K)$ is a finitely generated abelian group. The map $L_{P_0}$ is an essential tool to prove the finiteness of $X(K)$ for $g\ge2$.
\end{remark}


\begin{thm}[Abel]\begin{enumerate}
\item Let $D=\sum n_i(P_i)$ be a divisor on $X$, then $D=\div(f)$ for some $f\in\BC(X)^\times$ if and only if $\deg D=0$ and $\tr D=\oplus [n_i] \iota_{P_0}P_i=0\in J$.
\item We have an exact sequence
  \[0\to\{\div(f)\}\to \Div^0(X(\BC))\to J(\BC)\to 0.\]
\end{enumerate}\end{thm}

The proofs use Riemann $\theta$-function which replaces Weierstrass $\sigma$-function. Define
  \[\theta(z)=\sum_{n\in \BZ^g} \exp(i\pi^t nBn+2i \pi^t nz),\]
it converges because $\Im B$ is positive definite. If $u=a+Bb\in\Lambda$, $a,b\in\BZ^g$,
  \[\theta(z+u)=\theta(z)\exp(-i\pi^t bBb-2i\pi^tbz).\]
Hence the zeroes of $\theta$ are periodic of period $\Lambda$, and we can talk about the zeroes of $\theta$  in $J$, or $\theta\circ \iota_{P_0}$ in $X$.
\begin{thm}\begin{enumerate}
\item 
There is $w_0 \in \mathbb{C}^g$, unique up to
		$\Lambda$, such that if $z \in J$ is generic with a lifting $\tilde z \in \mathbb{C}^g$,
  \[\Function{\iota_P:B_g(0,r)}{Y(\BC)}{P}{\theta(w_0-\wt z+\iota_{P_0}(P))}\]
with $\iota_P(0)=P$
has divisor $(Q_{1,z})+\cdots+(Q_{g,z})$ where $Q_{1,z},\dots,Q_{g,z}$ are uniquely determined by
  \[\iota_{P_0}(Q_{1,z})\oplus \cdots\oplus \iota_{P_0}(Q_{g,z})=z\in J.\]
\item The map
  \[\Function{X^g/S_g}{J}{(P_1,\dots,P_g)}{\iota_{P_0}(P_1)+\dots+\iota_{P_0}(P_g)}\]
is a birational isomorphism.
\item The theta divisor $\Theta=\{x\in J: \theta(w_0-x)=0\}$ is
  \[\{\iota_{P_0}(Q_{1,z}),\dots,\iota_{P_0}(Q_{g,z}):Q_{i,z}\in X\}.\]
\end{enumerate}\end{thm}
\subsection{Differential forms}
Let $Y$ be a smooth algebraic variety over $\BC$ (we will take $Y=X$ or $J$), which is viewed as a complex analytic variety. By GAGA principal of Serre, the meromorphic functions on $Y(\BC)$ are one-to-one corresponding to rational functions  on $Y$.

If $\omega\in \Omega^1_{\BC(Y)}$, $P\in Y(\BC)$, then there is
  \[\iota_P:B(0,r)\ra Y(\BC)\]
with $\iota(0)=P$. Here $B_g(0,r)$ is the product of $g$ closed balls with radius $r$ of the complex plane. If $Y$ is of dimension $g$, we can write
  \[\iota_P^*\omega=f_1\diff z_1+\dots+f_g\diff z_g\]
for some meromorphic function $f_i$ on the open ball $B_g(0,1^-)$.

We say that $\omega$ is closed if locally, outside of the poles, it is $\diff f$. Then
  \[\iota_{P}^* \omega=\sum_{i=1}^g \frac{\partial f\circ \iota_P}{\partial z_i} \diff z_i.\]
By Poincar\'e's lemma, this is equivalent to $\diff \omega=0$, then
  \[0=\iota_P^* \diff \omega=\sum_{i=1}^g\diff  f_i\wedge \diff z_i=\sum_{i<j}\left(\frac{\partial f_i}{\partial z_j}-\frac{\partial f_j}{\partial z_i}\right) \diff z_j\wedge \diff z_i.\]
\begin{defn}
We say $\omega$ is of the
\begin{itemize}
\item {\em first kind}, if it is holomorphic and closed;
\item {\em second kind}, if locally $\omega=\diff f$ for some meromorphic $f$ (no residue);
\item {\em thrid kind}, if locally $\omega=\frac{\diff f}{f}$ for some nonzero everywhere $f$ (simple poles, integral residue).
\end{itemize}
\end{defn}
Then we have an exact sequence
  \[0\ra \RH^0(Y,\Omega^1)\ra \DSK(Y)\oplus \BC\otimes \DTK(Y)\ra (\Omega^1_{\BC(Y)})^{\diff=0}\ra 0.\]
Denote $\RH^1_\dR=\DSK(Y)/\{\diff f\}$, then we have a pairing (period)
  \[\Function{\RH^1_\dR(Y)\times\RH_1(Y(\BC),\BZ)}{\BC}{(\omega,u)}{\int_u \omega.}\]

We have several theorems similar to those for elliptic curves.
\begin{thm}\begin{enumerate}
\item $\iota_{P_0}^*$ induces an isomorphism  $\RH^1_\dR(J)\simeq \RH^1_\dR(X)$.
\item $\dim_{\BC}\RH^1_\dR(X)=2g$ and $(\omega,u)\mapsto\int_u\omega$ is perfect. Thus
  \[ \RH^1_\dR(X)=\Hom(\RH_1(X(\BC),\BZ),\BC).\]
\item If $u$ is generic, then the image of
  \[\eta_{i,u}=\diff\left(\frac{\partial\theta(z-u)/\partial z_i}{\theta(z-u)}\right)\in\DSK(J)\]
in $\RH^1_\dR(J)$ doesn't depend on $u$. Denote by $\eta_{i}=\iota_{P_0}^*\eta_{i,u}$, then $\omega_1,\dots,\omega_g,\eta_1,\dots,\eta_g$ is a basis of $\RH^1_\dR(X)$.
\item (Riemann period relation). If $u,v\in\RH_1(X(\BC),\BZ)$,
  \[\sum_{i=1}^g \int_u \eta_i\int_v \omega_i-\int_v\eta_i\int_u\omega_i=2\pi iu\#v.\]
\end{enumerate}\end{thm}
\begin{thm}[Theorem of square]
For any $\omega\in\DSK(J)$,
  \[m^*\omega-\pr_1^*\omega-\pr_2^*\omega=\diff f\]
for some $f\in\BC(J\times J)$.

For any $\omega\in\DTK(J)$,
  \[m^*\omega-\pr_1^*\omega-\pr_2^*\omega=\diff f/f\]
for some $f\in\BC(J\times J)^\times$.
\end{thm}
\begin{thm}
There is an algebraic group $\wt J$ with the following properties:
\begin{enumerate}
\item \[\wt J(\BC)=\frac{\DTK(X)}{\diff f/f}=\frac{\DTK(J)}{\diff f/f}=\BC^{2g}/\Lambda\] where $\Lambda$ is the lattice consisting of
  \[(\int_u \omega_1,\dots,\int_u\omega_g,\int_u\eta_1,\dots,\int_u\eta_g)\]
for all $u\in\RH_1(X(\BC),\BZ)$.
\item there is an exact sequence
  \[0\ra\RH^0(X,\Omega^1)\ra\wt J\sto{\pi} J\ra 0\]
with $\BC$-points
  \[0\ra \RH^0(X,\Omega^1)\ra \frac{\DTK(X)}{\{\diff f/f\}}\ra\frac{\Div^0(X)}{\{\div(f)\}}\ra0;\]
\item if $\eta\in \DSK(J)$, there is a unique $\alpha_\eta\in\RH^0(\wt J,\Omega^1)$, invariant under translation by $\wt J$, such that
  \[\pi^*\eta-\alpha_\eta=\diff f,\quad f\in\BC(\wt J). \]
$\RH^1_\dR(X)$ is isomorphic to the invariant forms on $\wt J$.
\end{enumerate}
\end{thm}


\section{$p$-adic fields}
\subsection{$p$-adic number}
Let $K$ be a field.
\begin{defn}
A {\em norm} on $K$ is a map $|\cdot|:K\ra\BR_+$ satisfying
\begin{itemize}
\item $|x|=0 \iff x=0$;
\item $|xy|=|x||y|$;
\item $|x+y|\le|x|+|y|$.
\end{itemize}
Say $|\cdot|$ is {\em ultrametric} or {\em non-archimedean} if $|x+y|\le \sup(|x|,|y|)$.

A {\em valuation} is a map $v:K\ra \BR\cup\{+\infty\}$ satisfying
\begin{itemize}
\item $v(x)=+\infty\iff x=0$;
\item $v(xy)=v(x)+v(y)$;
\item $v(x+y)\ge \inf(v(x),v(y))$.
\end{itemize}
Say $v$ is {\em discrete} if $v(K^\times)$ is discrete, i.e., $v(K^\times)=\alpha\BZ$  for some $\alpha>0$; {\em normalized} if $v(K^\times)=\BZ$.

$\pi$ is a {\em pseudo-uniformizer} If $v(\pi)>0$. If $v$ is discrete with $v(K^\times)=\alpha\BZ$, $\pi$ is a {\em uniformizer} if $v(\pi)=\alpha$.
\end{defn}
If $v$ is a valuation and $0< a<1$, then $x\mapsto |x|=a^{v(x)}$ is a norm. Conversely, if $|\cdot|$ is ultrametric, for any $\lambda>0$, $v(x)=-\lambda\log|x|$ is a valuation.

A norm or valuation defines a topology, in fact a metric space, with an open basis
  \[B(a,\delta^-)=\{x:|x-a|<\delta\}.\]

\begin{thm}[Ostrowski]\begin{enumerate}
\item On $\BQ$, up to equivalence, the nontrivial norms are $|\cdot|_\infty=|\cdot|_\BR$ and $|\cdot|_p=p^{-v_p(\cdot)}$.
\item On $\BC(T)$, up to equivalence, the nontrivial valuations are $v_a$, $a\in\BP^1(\BC)$.
\end{enumerate}\end{thm}
We have the product formula
  \[ \prod |x|_v=1, \quad x\in\BQ^\times;\]
  \[ \prod v_a(f)=0, \quad f\in\BC(T).\]
\begin{remark}
(1) If $|\cdot|$ is a ultrametric, $|\wh K|=|K|$ where $\wh K$ is the completion of $K$  under the topology induced by $|\cdot|$.

(2) If $(K,|\cdot|)$ is complete, $\sum a_n$ converges if and only if $a_n$ tends to $0$.

(3) Assume $K$ is complete. Let
  \[\CO_K=\{x\in K: |x|\le 1\}\]
be the ring of integers of $K$, then
  \[\CO_K\simeq\plim \CO_K/\{|x|\le a^n\}\]
for any $0<a<1$.
\end{remark}

Let $\BQ_p$ be the completion of $\BQ$ for $|\cdot|_p$ or $v_p$ and
  \[\BZ_p=\{x\in\BQ_p\mid |x|_p\le 1\}.\]
\begin{prop}
For any $n\ge 1$, $\BZ/p^n\BZ\simeq\BZ_p/p^n\BZ_p$.
\end{prop}
Thus $\BZ_p=\plim \BZ/p^n\BZ$.

Let $(K,v)$ be a complete field. Then all valuations on $K$ are equivalent and $K$ is complete for any of them.

For $s\ge 1$, let $P_s=K\oplus Kx\oplus\cdots\oplus Kx^{s-1}$. Let $g,h\in K[x]$ with $\deg g\le n,\deg h\le k$. Define
  \[\Function{\theta_{g,h}:P_k\oplus P_n}{P_{n+k}}{(u,v)}{ug+vh.}\]
Let $R=R(g,h)$ be the determinant of $\theta_{g,h}$. Then $R=0$ if and only if
  \[\deg g\le n-1,\ \deg h\le k-1\quad \text{or}\quad (g,h)\neq 1.\]
Denote
  \[v_0(\sum a_i x^i)=\inf_i v(a_i).\]
\begin{thm}[Hensel's lemma]\label{thm:hensel}
For $c>0, f,g,h\in\CO_K[x]$, suppose
\begin{itemize}
\item $\deg g\le n,\deg h\le k,\deg(f-gh)\le n+k-1$;
\item $v_0(f-gh)\ge c+2v(R(g,h))$.
\end{itemize}
Then there are unique $\wt g,\wt h$ with
\begin{itemize}
\item $\deg (g-\wt g),\le n-1,\deg(h-\wt h)\le k-1$;
\item $v_0(g-\wt g),v_0(h-h_0)\ge c+v(R(g,h))$;
\item $f=\wt g\wt h$.
\end{itemize}
\end{thm}
\begin{coro}
If $f\in K[x]$ is monic irreducible and $f(0)\in\CO_K$, then $f\in\CO_K[x]$.
\end{coro}
\begin{proof}
Write $f=x^d+a_{d-1}x^{d-1}+\dots+a_0$. Assume $i$ is the biggest one such that
  \[v(a_i)=\inf_j v(a_j)<0.\]
Then
  \[a_i^{-1}f=b_dx^d+\cdots+x^i+\dots+b_0,\quad b_i\in\CO_K.\]
Let $g=x^i+\cdots+b_0$ and $h=1+b_d x^{d-i}$. Then $R(g,h)\equiv 1\mod \fm_K$, where $\fm_K$ is the maximal ideal of $\CO_K$, and
  \[v_0(f-gh)>0,\quad\deg(f-gh)\le d-1.\]
Conclude the result by Theorem~\ref{thm:hensel}.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{thm:hensel}]
Write $\wt g=g+v, \wt h=h+u$, then we want
  \[f-gh-uv=gu+fv.\]
That is to say, $(u,v)$ is a fixed point of
  \[(u,v)\mapsto \theta_{g,h}^{-1}(f-gh-uv)=\varp(u,v).\]
It suffices to prove that $\varp$ is contracing on
  \[B=\{(u,v)\in P_k\oplus P_n: v_0(u,v)\ge \delta:=c+v(R)\}.\]
In fact,
  \[\begin{split}v_0(f-gh-uv)&\ge\inf (v_0(f-gh),v_0(uv))\\
  &\ge\inf (c+2\delta,2c+2\delta)=c+2\delta.\end{split}\]
Since $\theta^{-1}_{g,h}$ has entries in $R^{-1}\CO_K$, $v(\varphi(u,v))\ge c+2\delta-\delta=c+\delta$. Hence $\varphi(B)\subseteq B$.

For any $(u,v),(u',v')\in B$,
  \[\begin{split}
    &v_0(\varp(u,v)-\varp(u',v'))\\
    =&v_0(\theta^{-1}_{g,h}(u(v-v')+v'(u-u')))\\
    =&\inf(v_0(u)+v_0(v-v')-\delta,v_0(v')+v_0(u-u')-\delta)\\
    \ge & c+v_0(u-u',v-v'),
  \end{split}\]
thus $\varp$ is contracting.
\end{proof}
\begin{exam}
(1) If $f\in\CO_K[x]$, $\alpha\in\CO_K$ with $v(f(\alpha))>2v(f'(\alpha))$, then there is $\wt\alpha$ with $v(\wt\alpha-\alpha)>v(f'(\alpha))$ and $f(\wt\alpha)=0$.

(2) If $f\in\CO_K[x]$ is monic and $\alpha$ is a simple root of $f$ in the residue field $k_K$, then there is a unique lifting $\wt \alpha\in\CO_K$ with $f(\alpha)=0$.
\end{exam}

\begin{defn}
Let $V$ be a  vector space over $K$. A {\em valuation} on $V$ is a map $v:V\ra \BR\cup\{\infty\}$ satisfying
\begin{itemize}
\item $v(x)=+\infty\iff x=0$;
\item $v(\lambda x)=v(\lambda)+v(x)$;
\item $v(x+y)\ge \inf(v(x),v(y))$.
\end{itemize}
\end{defn}
\begin{thm}
Suppose $(K,v)$ is complete and $V$ is finite dimensional over $K$. Then all valuations on $V$ are equivalent and $V$ is complete for any one of them.
\end{thm}
\begin{proof}
Fix a basis $\{e_i\}$ of $V$. Define
  \[v_0(\sum x_i e_i)=\inf v(x_i).\]
Then
  \[v(\sum x_i e_i)\ge \inf_i (v(x_i)+v(e_i))\ge v_0(x)+inf_i v(e_i).\]
Suppose $v(\sum x_i^{(k)}e_i)$ tends to infinity but $\inf_i v(x_i^{(k)})$ tends to infinity. There is $c>0$ and $1\le i\le n$ such that $v(x_i^{(k)})\le c$ for any $k$, since $v((x_i^{(k)})^{-1}\sum x_i^{(k)}e_i)$ tends to infinity, $e_i$ lies in the closure of the space spanned by $e_1,\dots,e_{i-1},e_{i+1},\dots$.
\end{proof}
\begin{thm}
Suppose $(K,v)$ is complete and $L$ is a finite field extension of $K$, then there is a unique extension of $v$ as a field valuation on $L$:
  \[v(x)=\frac{1}{[L:K]}v(\rmN_{L/K}(x)).\]
\end{thm}
Let $G_K=\Gal(\ov K/K)$ be the absolute Galois group.
\begin{coro} \begin{enumerate}
\item $v$ extends uniquely to $\ov K$.
\item $G_K$ acts on $\ov K$ via isometrics $v(\sigma x)=v(x)$.
\item $G_K$ acts on $\wh{\ov K}$ continuously. Thus $G_K=\Aut(\wh{\ov K}/K)$.
\end{enumerate} \end{coro}

\begin{thm} \begin{enumerate}
\item $C=\wh{\ov K}$ is algebraic closed.
\item The residue field $k_C=k_{\ov K}=\ov k_K$.
\end{enumerate} \end{thm}

\subsection{No $2\pi i$ in $\BC_p$}
Let $\BC_p=\wh{\ov \BQ}_p$ be the completion of the algebraic closure of $\BQ_p$ with $v(\BC_p^\times)=v_p(\ov{\BQ}_p^\times)=\BQ$.  This field is non-canonically isomorphic to $\BC$ under assuming the Axiom of Choice.
We have an action of the Galois group $G_{\BQ_p}=\Gal(\ov\BQ_p/\BQ_p)=\Aut_{\cont}(\BC_p)$ on $\BC_p$.

\begin{thm}[Ax-Sen-Tate]
For any closed subgroup $H$ of $G_{\BQ_p}$, $\BC_p^H$ is the completion of $\ov{\BQ}_p^H$.
\end{thm}

Let $F$ be a field of characteristic zero with absolute Galois group $G_F=\Gal(\ov F/F)$. Let $\chi:G_F\ra \BZ_p^\times$ be the cyclotomic character, $\zeta_{p^n}\in\ov F$ be a primitive $p^n$-th root of unity. Then for any $\sigma\in G_F$, $\sigma(\zeta_{p^m})=\zeta_{p^m}^{\chi_m(\sigma)}$ with $\chi_m(\sigma)\in(\BZ/p^{m}\BZ)^\times$.

We have $\chi_m(\sigma\tau)=\chi_m(\sigma)\chi_m(\tau)$ and $\chi_m(\sigma)=\chi_{m-1}(\sigma)$ in $(\BZ/p^{m-1}\BZ)^\times$. Thus
  \[\chi(\sigma)=(\chi_m(\sigma))_{m\in\BN}\in\plim (\BZ/p^m\BZ)^\times=\BZ_p^\times,\]
and $\chi(\sigma\tau)=\chi(\sigma)\chi(\tau)$, $\sigma(\zeta)=\zeta^{\chi(\sigma)}$ for any $\zeta\in\mu_{p^\infty}$.

Now $2\pi i=p^n\log e^{\frac{2\pi i}{p^n}}$ and $\sigma(2\pi i)=p^n\log \zeta_{p^n}^{\chi(\sigma)}=\chi(\sigma)2\pi i.$
Tate proved that if $\sigma(x)=\chi(\sigma)x$ for any $\sigma\in G_{\BQ_p}$, then $x=0$.

\subsection{$p$-adic logarithm}
\begin{lemma}
If $v_p(x)>0$, then
  \[\log(1+x)=\sum_{n\ge 1} \frac{(-1)^{n-1}}{n}x^n\]
converges in $\BC_p$ and
  \[\log(1+x+y+xy)=\log(1+x)+\log(1+y),\quad v_p(x),v_p(y)>0.\]
\end{lemma}
\begin{proof}
Since $v_p(\frac{(-1)^n}{n}x^n)=nv_p(x)-v_p(n)\ge n v_p(x)-\frac{\log n}{\log p}$ tends to infinity as $n$ tens to infinity, the convergent is proved.
Since
  \[\log(1+X+Y+XY)=\log(1+X)+\log(1+Y)\]
holds as power series. Take $X=x$ and $Y=y$, then both sides are convergent.
\end{proof}
\begin{prop}
If $\CL\in\BC_p$, then there exists a unique $\log_\CL:\BC_p^\times\ra\BC_p$ satisfying
\begin{enumerate}
\item $\log_\CL(xy)=\log_\CL(x)+\log_\CL(y)$;
\item $\log_\CL(p)=\CL$;
\item $\log_\CL(x)=\sum_{n\ge 1}\frac{(-1)^{n-1}}{n}(x-1)^n$ if $v_p(x-1)>0$.
\end{enumerate}
\end{prop}
\begin{remark}
Choosing $\CL$ amounts to choosing a branch of $p$-adic logarithm.
Take $\CL=0$, we get Iwasawa logarithm $\log$. Then $\log_\CL x=\log x+\CL v_p(x)$.

For any $\sigma\in G_{\BQ_p}$, $\log \sigma(x)=\sigma(\log x)$ by unicity.

Also, we define
  \[\exp x=\sum_{n\ge 0}\frac{x^n}{n!},\]
which converges for $v_p(x)>\frac{1}{p-1}$.
\end{remark}
\begin{proof}
Choose $p^r$ for $r\in\BQ$ so that $p^{r+s}=p^r p^s$ (we only need to choose $p^{1/n!}$). Then for $x\in\BC_p^\times$, $x=p^{v_p(x)}y$ with $y\in\CO_{\BC_p}^\times$. Let $\bar y$ be its residue in $\ov\BF_p^\times=\CO_{\BC_p}/\fm_{\BC_p}$. Then there exists an integer $N$ such that $\bar y^N=1$ in $\ov\BF_p^\times$, i.e., $v_p(y^N-1)>0$. Define
  \[\log_\CL x=\CL v_p(x)+\frac{1}{N}\log y^N.\qedhere.\]
\end{proof}

\subsection{Cyclotomic extension}
For $n\ge 1$, let $F_n=\BQ_p(\zeta_{p^n})$.
\begin{prop}
$e_n=[F_n:\BQ_p]=(p-1)p^{n-1}$, $\pi_n=\zeta_{p^n}-1$ is a uniformizer of $F_n$ with $v_p(\pi_n)=\frac{1}{e_n}$ and $1,\zeta_{p^n},\ldots,\zeta_{p^n}^{e_n-1}$ is  a basis of $\CO_{F_n}$ over $\BZ_p$.
\end{prop}
\begin{proof}
The polynomial
  \[ \phi=\frac{(1+X)^{p^n}-1}{(1+X)^{p^{n-1}}-1}=X^{(p-1)p^{n-1}}+\cdots+p\]
kills $\pi_n$. Since $\phi$ is Eisenstein, $\phi$ is irreducible and $F_n=\BQ_p[X]/\phi$. Thus $e_n=(p-1)p^{n-1}$ and ${\bf N}_{F_n/\BQ_p}\pi_n=p$, this implies $v(\pi_n)=\frac{1}{e_n}v_p({\bf N}_{F_n/\BQ_p}\pi_n)=\frac{1}{e_n}$. And $v_p(F_n^\times)\subset \frac{1}{e_n} v_p(\BQ_p^\times)$, this implies that $\pi_n$ is a uniformizer.

Since $1,\pi_n,\ldots,\pi_n^{e_n-1}$ is a basis of $F_n$ over $\BQ_p$, for any $x\in F_n$,
  \[x=x_0+x_1 \pi_n+\cdots+ x_{e_n-1}\pi_n^{e_n-1}\]
for $x_i\in\BQ_p$. Notice that all nonzero terms have distinct valuation, thus $v_p(x)=\inf v_p(x_i \pi_n^i)$ and $v_p(x)\ge 0$ implies that $v_p(x_i)\ge 0$ for all $i$. Thus $1,\pi_n,\ldots,\pi_n^{e_n-1}$ forms a basis of $\CO_{F_n}$ over $\BZ_p$.
\end{proof}
\begin{cor}
Let $F_\infty=\cup F_n$, then $\chi:\Gal(F_\infty/\BQ_p)\simto \BZ_p^\times$.
\end{cor}
Define Tate's normalized trace map $R:F_\infty\ra \BQ_p$ as
  \[R(x)=\frac{1}{[F_n:\BQ_p]} \Tr_{F_n/\BQ_p} x,\quad x\in F_n.\]
\begin{prop}
$R$ extends by continuity to $\wh F_\infty\ra \BQ_p$ with \[R(\sigma(x))=R(x)=x\] for $x\in\BQ_p,\sigma\in\Gal(F_\infty/\BQ_p)$.
\end{prop}
\begin{proof}
We have $R(1)=1$,
  \[R(\zeta)=\begin{cases}
    -\frac{1}{p-1},\quad &\text{if}\ \zeta^p=1;\\
    0,\quad &\text{if}\ \zeta^p\neq1.
  \end{cases}\]
Thus $R(\CO_{F_n})\subseteq \BZ_p$ and $v_p(R(x))> v_p(x)-1$. This implies that $R$ is uniformly continuous and it can be extended to $\wh F_\infty$.
\end{proof}
\begin{thm}
For $k\in\BZ$ and $[K:\BQ_p]<\infty$,
  \[\BC_p(k)^{G_K}=\{x: \sigma(x)=\chi(\sigma)^k x,\forall \sigma\in G_{K}\}=\begin{cases}
    K,\ &\text{if}\ k=0;\\
    0,\ &\text{if}\ k\neq0.
  \end{cases}\]
\end{thm}
\begin{proof}
If $k=0$, this follows Ax-Sen-Tate. If $k\neq 0$, assume $0\neq x\in \BC_p(k)^{G_K}$, $y=\log x$, $\sigma(y)=y+k\log \chi(\sigma)$ for any $\sigma$. By Ax-Sen-Tate, $y\in\wh F_\infty=(\ov\BQ_p^{\ker \chi})\wh{\ }$. Then $R(\sigma(y))=R(y)+k\log \chi(\sigma)$. But $R(y)\in\BQ_p$, $\sigma(R(y))=R(y)$, ridiculous!
\end{proof}

\section{Fontaine's rings and $p$-adic Galois representations}
\subsection{$p$-rings}
\begin{defn}
Let $A$ be a ring and $I$ be an ideal. Say $A$ is \emph{separated and complete for $I$-adic topology} if $A\simto \plim (A/I^n)$. In this case, the $I$-adic topology on $A$ and discrete topology on $A/I^n$ turns this into an isomorphism of $I$-adic topology rings.
\end{defn}
In this case, $\sum x_n$ converges iff $x_n\ra 0$, i.e., for any $N$, there exists $n_0$ such that $x_n \in I^N$ for $n\ge n_0$.
\begin{example}
If $(K,v)$ is complete, $v(\pi)>0$, then $\CO_K$ is separated and complete for $\pi$-adic topology.
\end{example}

\begin{lemma}
Assume $A$ is separated and complete for $\pi$-adic topology, $\pi $ is not a zero divisor, $S$ a system of representatives of $A/\pi$ inside $A$. Then any $x\in A$ can be written as $x=\sum_{i\ge 0} s_i \pi^i$ with $s_i\in S$ uniquely.
\end{lemma}
\begin{proof}
There is a unique $s(x)\in S$ such that $x-s(x)\in \pi A$. Let $x_0=x,x_n=\frac{1}{\pi}(x_{n-1}-s(x_{n-1})$, then
  \[x=\sum_{i=0}^n s(x_i) \pi^i+\pi^{n+1} x_{n+1}.\]
Take $s_i=s(x_i)$.
\end{proof}
\begin{defn}
Let $R$ be a ring of characteristic $p$. $R$ is called \emph{perfect} if $x\mapsto x^p$ is an isomorphism. $I$ is \emph{perfect} if $R/I$ is perfect, i.e., $x\mapsto x^p$ is bijective on $I$.

$A$ is called a \emph{$p$-ring} with residue ring $R$ if there is $\pi$ such that $A$ is separated and complete for $\pi$-adic topology and $A/\pi=R$, in particular, $p\in\pi A$. $A$ is \emph{strict} if $pA=\pi A$. $A$ is  \emph{perfect} if strict and $R$ is perfect.
\end{defn}
\begin{example}
\begin{enumerate}
\item $\BZ_p$ is perfect.
\item Let $J$ be a set and $W_J=\BZ_p[X_j^{p^{-\infty}},j\in J]$, then
  \[\wh W_J=\plim W_J/p^n W_J\]
is a perfect ring with residue ring $\ov{W}_J=\BF_p[X_j^{p^{-\infty}},j\in J]$.
\end{enumerate}
\end{example}
If $A$ is perfect, then $A/p$ is perfect. If $R$ is perfect, there is a unique perfect $A$ with $A/p=R$.

\subsection{Teichm\"uller representatives}
Let $A$ be a $p$-ring and $R=A/\pi$.
\begin{lemma}
If $x-y\in\pi A$, then $x^{p^n}-y^{p^n}\in \pi^{n+1}A$.
\end{lemma}
\begin{proof}
By induction.
\end{proof}
For any ring $S$, Denote
  \[\fR(S)=\{x=(x^{(n)})_{n\in\BN}: x^{(n)}\in S, (x^{(n+1)})^p=x^{(n)}\}.\]
\begin{prop}
We have $\fR(A)=\fR(R)$. If $x=(x^{(n)})\in \fR(R)$, let $\hat x^{(n)}\in A$ be a  lifting of $x^{(n)}$, then $(\hat x^{(n+k)})^{p^k}$ tends to $\tilde x^{(n)}\in A$ and $\tilde x=(\tilde x^{(n)})\in \fR(A)$.
\end{prop}
\begin{cor}
$\fR(A)$ is a ring with ring structure as $\fR(R)$, which is a perfect ring of characteristic $p$.
\end{cor}
This is an old construction of Fontaine. Scholze calls it the \emph{tilt} $A^\flat$ of $A$.
\begin{example}
$\BZ_p^\flat=\fR(\BF_p)=\BF_p$. More generally, $A^\flat=A/p$ if $A$ is perfect, because if $R$ is perfect, $\fR(R)=R$.
\end{example}
\begin{remark}
(1) If $x\in R$, then $x=(x,x^{1/p},\ldots)\in \fR(R)$ gives $\tilde x\in\fR(A)$. Then $[x]=\tilde x^{(0)}$ is called the \emph{Teichm\"uller lifting} of $x$, it's the unique lift to $A$  of $x$ with $p^n$-th root, for any n. We have
  \[[x]=\lim_{n\ra+\infty} (\wh{x^{1/p^n}})^{p^n}.\]
and $[xy]=[x][y]$.

(2) If $A$ is strict, any $x\in A$ can be written as $\sum_{x\ge 0} [x_i]p^i$ for $x_i\in R$.
\end{remark}

A question is: can we write $+$ and $\times$ in $A$ using this decomposition? The answer is yes, and the tool is Witt vector.
\begin{thm}
(1) Assume $R$ is a perfect ring of characteristic $p$. There is a unique strict $p$-ring $W(R)$ unique up to unique isomorphism such that $W(R)/p=R$.

(2) If $A$ is a $p$-ring, $A/\pi=R'$, $\bar \theta:R\ra R'$, $\tilde \theta:R\ra A$ with $\tilde\theta(xy)=\tilde\theta(x)\tilde\theta(y)$, then there is a unique ring morphism $\theta:W(R)\ra A$ lifting $\bar \theta$ such that $\theta([x])=\tilde \theta(x)$.
\end{thm}

\begin{remark}
(1) The unicity in (2) is obvious, for $x=\sum [x_i]p^i\in W(R)$, $\theta(x)=\sum p^i\tilde \theta (x_i)$. $W(R)$ is unique since there is a unique $\theta:W(R)\ra W(R)$ identity modulo $p$ for $\bar \theta(x)=x$ and $\tilde \theta(x)=[x]$. There is a unique lifting of $x$ with $p^n$-th roots for any $n$, namely $[x]$, thus $\theta=\id$.

(2) If $R'$ is perfect, $\Hom(W(R),W(R'))=\Hom(R,R')$ for $\tilde\theta(x)=[\bar\theta(x)]$.

The Frobenius $\varphi:W(R)\ra W(R)$ is the lifting of $x\mapsto x^p$, i.e.,
  \[\varphi(\sum [x_i]p^i)=\sum[x_i^p]p^i.\]

(3) If $A$ is perfect, then $W(A/p)=A$. In particular, $W(\BF_p)=\BZ_p$ and $W(\ov W_J)=\wh W_J$.
\end{remark}
Now we prove that $\wh W_J$ satisfies (2). The map $f:W_J\ra A, f(x_j^{p^{-n}})=\tilde{\theta}(x_j^{p^{-n}})$ by continuity extends $f$  to $\hat f:\wh W_J\ra A$ (provides $A$ is $p$-adically complete). We will show $\hat f([x])=\tilde \theta(x)$ for any $x\in\ov W_J$. Since $\hat f$ modulo $\pi$ is $\bar \theta, \hat f([x])-\tilde\theta(x)\in\pi A$, thus
  \[\hat f([x^{p^{-n}}])-\tilde\theta(x^{p^{-n}})\in \pi A \]
and then $\hat f([x])-\tilde\theta(x)\in\pi^{n+1} A$. In general, $R$  can be written as $\ov W_J/I$ for some perfect ideal $I$.
Let
\[W(I)=\{\sum p^i[x_i]: x_i\in I\}\subset \wh W_J.\]
\begin{lemma}\label{lem:WI}
$W(I)$ is an ideal of $\wh W_J$ and we take $W(R)=\wh W_J/W(I)$.
\end{lemma}
Let $U=\BN\sqcup \BN=\{1,2\}\times \BN$ and $\Sigma(X)=\sum [X_i]p^i,\Sigma(Y)=\sum [Y_i]p^i \in\wh W_U$, then
\[\begin{split}\Sigma(X)+\Sigma(Y)&=\sum [s_i(X,Y)] p^i\\
\Sigma(X)\Sigma(Y)&=\sum [p_i(X,Y)] p^i\end{split}\]
for $s_i,p_i\in \ov W_U$.
\begin{prop}
Let $A$ be a perfect $p$-ring with $A/p=R$. For $x=(x_i), x_i\in R$, let $\Sigma(x)=\sum [x_i]p^i\in A$. Then
\[\begin{split}\Sigma(x)+\Sigma(y)&=\sum [s_i(x,y)] p^i\\
\Sigma(x)\Sigma(y)&=\sum [p_i(x,y)] p^i.\end{split}\]
\end{prop}
\begin{proof}
Let $\bar\theta:\ov W_U\ra R, \bar\theta(X_i)=x_i, \bar\theta(Y_i)=y_i$ and $\tilde \theta:\ov W_U\ra A, \tilde\theta(x)=[\bar\theta(x)]$, then there is a unique $\theta:\wh W_U\ra A$ with $\theta([x])=[\bar\theta(x)]$. Now
\[\begin{split}&\Sigma(x)+\Sigma(y)=\theta(\Sigma(x))+\theta(\Sigma(y))=\theta(\Sigma(x)+\Sigma(y))\\
=&\theta(\sum [s_i(x,y)]pi)=\sum p^i [\bar\theta(s_i(x,y))]=\sum p^i [s_i(x,y)]. \end{split}\]
Similar for product.
\end{proof}
\begin{proof}[Proof of Lemma~\ref{lem:WI}]
$\Sigma(0)=0$ implies that $S_i$ has no constant term and $W(I)$ is stable under addition. $\Sigma(x)=\Sigma(y)=0$ if $x=0$ or $y=0$ implies $p_i$ has no  term of degree $0$ in $X$ or $Y$. This implies that $W(I)$ is stable by multiplication by $\wh W_J$.
\end{proof}

\subsection{The ring $\wt E^+$}
$\fR(A)$ is a perfect ring of characteristic $p$. Define $\wt E^+=\fR(\CO_{\BC_p})=\fR(\CO_{\BC_p}/p)$ (i.e., Fontaine's $R$ or Scholze's $\CO_{\BC_p^\flat}$).
The Galois group $G_{\BQ_p}$ acts via the action on every component.

If $x=(x^{(n)})\in \wt E^+$, let $x^\sharp=x^{(0)}$, then $(xy)^\sharp=x^\sharp y^\sharp$. Let $v_E(x)=v_p(x^\sharp)$.
\begin{thm}
\begin{enumerate}
\item $\wt E^+$ is a perfect ring of characteristic $p$, $v_E$  is a valuation on $\wt E^+$ for which it is complete.
\item $G_{\BQ_p}$ acts continuously, compatible with  ring structure, commutes with $x\mapsto x^p$.
\item $\wt E:=\Fr \wt E^+=\wt E^+[\frac{1}{\varpi}]$ for any $\varpi$ with $v_E(\varpi)>0$  is algebraically closed.
\end{enumerate}
\end{thm}
\begin{proof}
(1) One can check that $v_E$ is a valuation directly. If $v_E(x-y)\ge p^m$, then $v_E(x^{1/p^m}-y^{1/p^m})\ge 1$ and $v_p(x^{(m)}-y^{(m)})\ge 1$, i.e., $x^{(m)}=y^{(m)}$ in $\CO_{\BC_p}/p$. Thus $x^{(i)}=y^{(i)}$ in $\CO_{\BC_p}/p$ for $i\le m$. Since the topology of  $\wt E^+$ is induced by the product topology of discrete topology on $\CO_{\BC_p}/p$, $\wt E^+$ is complete for $v_E$.

(2) $G_{\BQ_p}$ respects the ring structure obvious.
Since $v_E(\sigma(x))=v_p(\sigma(x^\sharp))=v_p(x^\sharp)=v_E(x)$, $G_{\BQ_p}$ acts by isometries.

Let $M\ge 0$, choose $p^n\ge M$, $y\in \CO_{\ov \BQ_p}$ with $v_p(y-x^{(n)})\ge 1$. There is a finite Galois extension $K/\BQ_p$ with $y\in K$. For $\sigma\in G_{\BQ_p}$ and $\tau\in G_K$,
\[\sigma\tau(x^{(n)})-\sigma(x^{(n)})=\sigma\tau(x^{(n)}-y)-\sigma(x^{(n)}-y)\]
has valuation $\ge 1$, thus $v_E(\sigma\tau(x)-\sigma(x))\ge p^n\ge M$, i.e., $\sigma\mapsto\sigma(x)$ is continuous.

(3) It's enough to prove that for any unitary $P$ in $\wt E^+[X]$ has a root in $\wt E^+$. Let $P=Q^{p^k}$ with $Q'\neq 0$. We may assume $(P,P')=1$, then there exist $U,V\in\wt E^+[X]$, $UP+VP'=\varpi$ for some $\varpi\in \wt E^+$ with $v_E(\varpi)> 0$.

Write $P(X)=X^d+a_{d-1}X^{d-1}+\cdots+a_0$ with $a_i=(a_i^{(n)})$. Choose $p^N > 2 v_E(\varpi)$. Choose $(x^{(n)})\in\wt E^+$ such that $P^{(i)}(x^{(N)})=0$ where $P^{(i)}(X)=X^d+a_{d-1}^{(i)}X^{d-1}+\cdots+a_0^{(i)}\in\CO_{\BC_p}[x]$. Then $P(x)^{(N)}=0$ in $\CO_{\BC_p}/p$, thus
\[v_E(P(x))\ge p^N>2 v_E(\varpi)\ge 2 v_E(P'(x)).\]
By Hensel's lemma, $P$ has a root $y$ with $v_E(y-x)\ge v_E(P(x))-v_E(P'(x))$.
\end{proof}

Fix $\vare=(1,\vare^{(1)},\ldots)\in \wt E^+$ with $\vare^{(1)}\neq 1$. Then $\vare^{(n)}$ is a primitive $p^n$-th root of unity and $$v_E(\vare-1)=\lim_{n\to+\infty}p^nv_p(\vare^{(n)}-1)=\frac{p}{p-1}>0.$$
\begin{prop}
If $\sigma\in G_{\BQ_p}$, $\sigma(\vare)=\vare^{\chi(\sigma)}=\sum {\chi(\sigma)\choose i} (\vare -1)^i$.
\end{prop}
If $x\in \CO_{\BC_p}$, note by $x^\flat$ any element of $\wt E^+$ with  $(x^\flat)^\sharp=x$. Note that $x^\flat$ is only unique up to $\vare^{\BZ_p}$.

Since $v_E(\vare-1)>0$, $E_{\BQ_p}=\BF_p((\vare-1))\inj \wt E$ implies $E=E_{\BQ_p}^{\sep}\inj\wt E$.
\begin{thm}[Fontaine-Wintenberger]
\begin{enumerate}
\item $\wt E$ is the completion of $E$ for $v_E$. If $\CH=\ker \chi$, then $\CH$ acts trivially on $E_{\BQ_p}$ and $\CH\inj \Gal(E/E_{\BQ_p})$.
\item $\CH\simeq \Gal(E/E_{\BQ_p})$.
\end{enumerate}
\end{thm}
\begin{remark}
We get a d\'eversage
\[1\ra G_{\BF_p((T))}\ra G_{\BQ_p}\sto{\chi}\BZ_p^\times\ra 1.\]
This is very useful to study $G_{\BQ_p}$ and its representations.
\end{remark}

\subsection{The ring $\wt A^+=W(\wt E^+)$}
Any $x\in\wt A^+$ can be written uniquely as $\sum [x_i]p^i$ for $x_i\in\wt E^+$. It commutes with $G_{\BQ_p}$-action and $\varp$-action.
\begin{thm}
\begin{enumerate}
\item $\theta:\wt A^+\ra \CO_{\BC_p}$, $\theta(\sum[x_i] p^i)=\sum p^i x_i^\sharp$ is a surjective ring morphism commuting with $G_{\BQ_p}$-actions.
\item $\ker\theta$ is principal and $x\in \ker \theta$ is a generator if and only if $v_E(x_0)=1$.
\end{enumerate}
\end{thm}
\begin{proof}
(1) $\bar \theta:\wt E^+\ra \CO_{\BC_p}/p$ and $\tilde \theta:\wt E^+\ra \CO_{\BC_p}, \tilde\theta(x)=x^\sharp$ give the unique $\theta$ with $\theta([x])=x^\sharp$.

(2) Define $\bar x=x_0$ if $x=\sum [x_i]p^i$. If $\theta(x)=0$, then $x_0^\sharp=-\sum_{i\ge 1} p^i x_i^\sharp$, thus $v_p(x_0^\sharp)\ge 1$  and $v_E(x_0)\ge 1$. If $\theta(x)=\theta(y)=0$ and $v_E(\bar x)=1$, $v_E(\bar y)\ge 1$, then there is $a_0\in\wt E^+$ such that $\bar y=\bar x a_0$, $y=x[a_0]+py_1$ with $\theta(y_1)=0$. Thus  $y=x(\sum [a_i]p^i)$.
\end{proof}
For example, $[p^\flat]-p$ and
\[\omega=\frac{[\vare]-1}{[\vare^{1/p}]-1}\]
are two different generators of $\ker \theta$.

The natural topology on $\wt A^+$ is $(p,[p^\flat])=(p,\ker\theta)$-adic topology, and on $\wt E^+$ is $v_E$ or $p^\flat$-adic topology. Then $\wt A^+\ra\wt E^+$ is continuous for the natural topology and the natural topology turns the bijection $(\wt E^+)^\BN\ra \wt A^+$ into a homeomorphism.
The basis for open sets are $x+p^n\wt A^++\omega^{k-1}\wt A^+$ for $n,k\in\BN$. The action of $G_{\BQ_p}$ is  continuous under this topology (but not for the $p$-adic topology).

We have
  \[\sigma([\vare])=[\sigma(\vare)]=[\vare^{\chi(\sigma)}]=[\vare]^{\chi(\sigma)}=\sum_{k=0}^{+\infty}{\chi(\sigma)\choose k}([\vare]-1)^k.\]


\subsection{The ring $B_{\dR}^+$ and the field $B_\dR$}

We extend  $\theta$ to $\wt A^+[\frac{1}{p}] \ra \BC_p$, it's still a ring morphism with kernel generated by $\omega$. Let $B_\dR^+$ be the completion of $\wt A^+[\frac{1}{p}]$ for the $(\ker \theta)$-adic topology, i.e.,
  \[B_\dR^+=\plim \wt A^+[\frac{1}{p}]/(\ker\theta)^k.\]
This is a complete discrete valued ring with residue field $\BC_p$. The valuation $v_H$ is normalized by  $v_H(\omega)=1$. Since $\theta$ commutes with the action of $G_{\BQ_p}$, $\ker\theta$ is stable by $G_{\BQ_p}$ and $G_{\BQ_p}$ acts on $B_\dR^+$.

Then natural topology on $B_\dR^+$ is defined as follows: the basis of open sets are $x+p^n\wt A^++\omega^{k+1}B_\dR^+$. This is the projective limit topology, each $B_\dR^+/(\ker\theta)^k$  endowed with the $x+p^n\wt A^+$ as a basis  of open sets. $B_\dR^+$ is a Fr\'echet space as a projective limit of Banach spaces. The $G_{\BQ_p}$-action is continuous.

\begin{lemma}
If $x\in B_\dR^+$, $v_p(\theta(x))>0$, then
  \[\log(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}x^n\]
converges in $B_\dR^+$ and
  \[\log(1+\sigma(x))=\sigma(\log(1+x)).\]
\end{lemma}
\begin{proof}
Choose $a\in\BN$ with $av_p(\theta(x))\ge 1$, then $x^a\in p\wt A^++\omega B_\dR^+$. Write $x^a=pu+\omega v$ and $n=aq+r$ with $0\le r< a-1$. Assume $v\in p^{-N_k}\wt A^++\omega^{k+1} B_\dR^+$, then
  \[x^n=x^r(x^a)^q=x^r(pu+\omega v)^q\in p^{q-kN_k} \wt A^++\omega^{k+1} B_\dR^+. \]
Since $q$ is nearly $n/a$, $x^n/n$ tends to zero modulo $\ker\theta$.
\end{proof}

Now
  \[t=\log[\vare]=\sum_{n=1}^{+\infty} \frac{(-1)^{n-1}}{n} ([\vare]-1)^n \]
converges in $B_\dR^+$ since $v_p(\theta([\vare]-1))>0$. And
  \[\sigma(t)=\log \sigma([\vare])=\log[\vare]^{\chi(\sigma)}=\chi(\sigma)\log[\vare]=\chi(\sigma) t,\]
that is to say, $t$ is the $p$-adic analogy of $2\pi i$.

\begin{prop}
$t$ is a generator of $\ker \theta$, in particular, $t\neq 0$.
\end{prop}
\begin{proof}
Since $[\vare]-1=\omega([\vare^{1/p}]-1)$,
  \[\theta(\frac{t}{\omega})=\theta(\frac{t}{[\vare]-1})\theta([\vare^{1/p}]-1)\neq 0.\qedhere\]
\end{proof}

Let $B_\dR=B_\dR^+[\frac{1}{t}]$ be the fraction field of $B_\dR^+$. We extend the action of $G_{\BQ_p}$ by $\sigma(\frac{1}{t})=\frac{1}{\chi(\sigma)t}$.
\begin{thm}
(1) $\bar\BQ_p$ is a subfield of $B_\dR^+$. More precisely, $\theta$ induces an isomorphism for the separable closure of $\BQ_p$ inside $B_\dR^+$ to $\ov \BQ_p$.

(2) If $[K:\BQ_p]<\infty$, $(B_\dR)^{G_K}=K$.
\end{thm}

\begin{proof}
(1) Let $P\in\BQ_p[X]$ be the minimal polynomial of $x\in \ov \BQ_p$ with $(P,P')=1$. Let $\hat x\in B_\dR^+$ satisfy $\theta(\hat x)=x$, then
$v_H(P(\hat x))\ge 1$ and $v_H(P'(\hat x))=0$. By Hensel's lemma, $P$ has a unique root in $\hat x+\omega B_\dR^+$.

(2) If $x\in B_\dR^{G_K}-\{0\}$, write $x=t^k y$ with $y\in B_\dR^+$ and $\theta(y)\neq 0$. Then
  \[\sigma(\theta(y))=\chi(\sigma)^{-k}\theta(y).\]
by Tate's lemma, $k=0$ and $\theta(y)\in K$, and then $x-\theta(x)$ is fixed by $G_K$ with $v_H>0$. Finally $x=\theta(x)\in K$.
\end{proof}
\begin{remark}
(1) Can the inclusion $\ov \BQ_p \inj  B_\dR^+$ extend to $\BC_p$ continuously? No, because $\ov \BQ_p$ is dense in $B_\dR^+$.

(2) By Ax-Sen-Tate, $t$ is not in the closure of $\BQ_p(\mu_{p^\infty})$ in $B_\dR^+$.
\end{remark}

Define a sequence of sub-rings of $\bar\BQ_p$,
  \[\CO^{(0)}=\CO_{\bar\BQ_p},\quad \CO^{(k+1)}=\ker(\CO^{(k)}\ra \CO^{(k)}\otimes\Omega^1_{\CO^{(k)}/\BZ_p}).\]
They have a basis of open subsets $x+p^n\CO^{(k)}$ and
  \[B_\dR^+=\plim_k (\plim_n (\CO^{(k)}/p^n \CO^{(k)})[\frac{1}{p}]).\]

\subsection{$p$-adic Galois representation}
Let $K$ be a finite extension of $\BQ_p$ and $G_K=\Gal(\ov \BQ_p/K)$. A $\BQ_p$-representation of $G_K$ is a finite dimensional $\BQ_p$-vector space $V$ endowed with a continuous linear action of $G_K$.

If $\dim V=d$ with basis $e_1,\ldots,e_d$, let $U_\sigma=(a_{i,j})$ be the matrix of $\sigma$, then $\sigma\mapsto U_\sigma$ is a continuous group homomorphism $G_K\ra \GL_d(\BQ_p)$, where $1+p^n M_d(\BZ_p)$ is a basis of open subgroups of $\GL_d(\BQ_p)$.

\begin{example}
(1) $k\in\BZ$, $V=\BQ_p(k)=\BQ_p e(k)$, where $\sigma(e(k))=\chi(\sigma)^k e(k)$.

(2) Let $E/K$  be an elliptic curve, then $G_K$ acts on $E(\ov K)[p^n]\simeq(\BZ/p^n\BZ)^2$ continuously. Let
  \[T_p(E)=\plim_n E(ov K)[p^n]\]
be the Tate module, then $T_p(E)$ is a $\BZ_p$-module of rank $2$ with continuous $G_K$-action. In fact, $T_p(E)=\BZ_p\otimes \RH_1(E(\BC),\BZ)$.
Let $V_p(E)=\BQ_p\otimes T_p(E)$, this is a $\BQ_p$-representation of dimension $2$.

(3) Let $X/K$ be a curve of genus $g$ with Jacobian $J$ and $V_p(J)=T_p(J)\otimes \BQ_p$, this is a $\BQ_p$-representation of dimension $2g$.

(4) $\RH^1_\et(X_{\ov K},\BQ(k))$ is a $\BQ_p$-representation of $G_K$ if $X$ is an algebraic variety defined over $K$.

(5) Let $V$ be a $\BQ_p$-representation, then $V^*=\Hom(V,\BQ_p)$ is also a $\BQ_p$-representation under $\sigma.\ell(v)=\ell(\sigma^{-1}. v)$ and the matrix is ${}^t U_\sigma^{-1}$ under the dual basis.
\end{example}

To study $\BQ_p$-representation of $G_K$, there is a very fruitful strategy of Fontaine.
\begin{itemize}
\item define rings $B$ with an action of $G_K$ with extra structures stable by $G_K$, e.g., $B=B_\dR$ and $\Fil^i B_\dR=t^i B_\dR^+, i\in\BZ$.
\item $D_B(V)=(B\otimes V)^{G_K}$ and $D_B^*=\Hom_{G_K}(V,B)=(B\otimes V^*)^{G_K}$ are $B^{G_K}$-modules ($B^{G_K}$ is a ring) with extra structures.
\end{itemize}
The art is to construct interesting $B$'s, Fontaine is a master: $B_\dR^+,B_\dR,B_\cris,B_\st$.
\begin{example}
$D_{\dR}(V)=(B_{\dR}\otimes V)^{G_K}$ is a $K$-vector space with filtrations.
\end{example}

If $e_1,\ldots,e_d$ is a basis of $B\otimes V$ over $B$, $U_\sigma$ is the matrix of $\sigma$, then $U_{\sigma\tau}=U_\sigma \sigma(U_\tau)$. Say that $V$ is $B$-admissible if there is a basis in which $U_\sigma=1$ for all $\sigma.$ If you start from any $U_\sigma$, that's equivalent to say, there exists $M\in \GL_d(B)$ such that $U_\sigma \sigma(M)=M$.
\begin{prop}
If $B$ is a field, $B^{G_K}$ is a field and $\dim_{B^{G_K}} D_B(V)\le \dim V$ with equality iff $V$ is  $B$-admissible.
\end{prop}
\begin{proof}
Let $x_1,\ldots,x_r\in D_B(V)\subset B\otimes V$ dependent over $B$. Assume $\lambda_1 x_1+\cdots+\lambda_r x_r=0$, take a minimal one and $\lambda_1=1$. Then
  \[x_1+\sigma(\lambda_2)x_2+\cdots+\sigma(\lambda_r)x_r=0\]
and
  \[(\sigma(\lambda_2)-\lambda_2)x_2+\cdots+(\sigma(\lambda_r)-\lambda_r)x_r=0.\]
By minimality, $\sigma(\lambda_i)=\lambda_i$  and $\lambda_i\in B^{G_K}$. Thus
  \[\dim_{B^{G_K}} D_B(V)\le \dim_B (B\text{-space generated by }D_B(V))\le \dim V.\]
The equality holds iff there is a basis of $B\otimes V$ with elements in $D_B(V)$, i.e., $V$ is $B$-admissible.
\end{proof}
\begin{prop}
$V$ is $B$-admissible iff $V^*$ is also $B$-admissible.
\end{prop}
\begin{proof}
That's because if $U_\sigma\sigma(M)=M$, then ${}^tU_\sigma^{-1}\sigma({}^tM^{-1})={}^tM^{-1}$.
\end{proof}
\begin{prop}
$V$ is $\ov\BQ_p$-admissible iff $G_K$ acts through a finite quotient.
\end{prop}
\begin{proof}
$\Rightarrow$: $U_\sigma=M\sigma(M)^{-1}$ for some $M\in \GL_d(L)$ with $L/\BQ_p$ finite Galois.

$\Leftarrow$: Pick such $L$ with $H=\Gal(L/\BQ_p)$, then for any $\alpha\in L$, let $M=\sum_{\tau\in H}\tau(\alpha) U_\tau$, then
  \[U_\sigma \sigma(M)=\sum_{\tau\in H} U_\sigma \sigma\tau(\alpha)V_\tau=\sum_{\tau\in H} \sigma\tau(\alpha)U_{\sigma\tau}=M.\]
We want $\det M\neq 0$. $\det(\sum X_\tau U_\tau)=\sum X_\tau^d\det U_\tau+\cdots$, it's nonzero because Arthur's independence of characters.
\end{proof}

\begin{thm}
(1) $\BQ_p(k)$ is $\BC_p$-admissible iff $k=0$ (Tate's theorem).

(2) $V$ is $\BC_p$-admissible iff $I_K$ acts through a finite quotient where
  \[0\ra I_K\ra G_K\ra \Gal(\ov \BF_p/k_K)\ra 1.\]
\end{thm}
\begin{remark}
(1) $\BQ_p(k)$ is  $B_\dR$-admissible (=de Rham), thanks to $t^{-k}$.

(2) Fontaine conjectures that $\RH^1_\et(X_{\bar K},\BQ_p(k))$ are de Rham.

(3) We are going to prove $V_p(J)$ is de Rham if $J$ is the Jacobian of curve $X/K$.
\end{remark}

\section{$p$-adic abelian integral}
\subsection{Lubin-Tate formal groups}
Assume $h=[K:\BQ_p]<\infty$, $k_K=\BF_q$, $q=p^f$, $\pi$ is a uniformizer of $K$. Since $x^q=x$ in $\BF_q$, $x^q-x\in \pi\CO_K$ for $x\in \CO_K$. Then
$\CO_K\supset W(k_K)$ and $\CO_K=W(k_K)[x]/(\phi)$ for an Eisenstein polynomial $\phi$.
$K_0=W(k_K)[\frac{1}{p}]$ is the maximal unramified subfield of $K$, and $K/K_0$ is totally ramified of degree $e=\deg \phi$ where $h=ef$. Let $P$ be a polynomial with \[P\equiv\pi X+X^q\mod \pi X^2\CO_K[[X]].\]
\begin{lemma}
If $a_1,\ldots,a_d\in\CO_K$ and $\ell=a_1X_1+\cdots+a_dX_d$, then there is a unique $F_\ell\in \ell+I^2$ where $I=(X_1,\ldots,X_d)\subset \Lambda=\CO_K[[X_1,\ldots,X_d]]$, such that
  \[P(F_\ell(X_1,\ldots,X_d))=F_\ell(P(X_1),\ldots,P(X_d)).\]
\end{lemma}
\begin{proof}
We will construct $F_n\in\Lambda$ such that $F_{n+1}-F_n\in I^{n+1}$ and $P(F_{n})-F_n(P)\in \pi I^{n+1}$, then we can take $F_1=\ell$ and $F_\ell=\lim F_n$.
We have
\[\begin{split}
P(\ell)&=\pi \ell+\ell^q\equiv \pi\ell+\sum_{i=1}^d a_i^q X_i^q\mod \pi I^2,\\
\ell(P)&=\pi\ell+\sum_{i=1}^d a_i X^q_i,\\
P(\ell)-\ell(P)&\equiv\sum (a_i^q-a_i)X_i^q\equiv 0\mod \pi I^2.\end{split}\]
Assume $F_{n+1}=F_n+R_n$ where $R_n$ is homogeneous of degree $n+1$, then
\[\begin{split}P(F_{n+1})&\equiv P(F_n)+\pi R_n+R_n^q\mod \pi I^{n+1}\\
F_{n+1}(P)&\equiv F_n(P)+\pi^{n+1} R_n+R_n(X^q)\mod \pi I^{n+1}\end{split}\]
Take $R_n=\frac{(P(F_n)-F_n(P))^{n+1}}{\pi^{n+1}-\pi}\in\CO_K[[X_1,\ldots,X_d]]$, then
  \[P(F_{n+1})-F_{n+1}(P)\equiv R_n(X)^q-R_n(X^q)\equiv 0 \mod \pi I^{n+1}.\qedhere\]
\end{proof}
Denote
  \[X\oplus Y=F_{X+Y}\in\CO_K[[X,Y]],\]
then
  \[P(X)\oplus P(Y)=P(X\oplus Y)\]
and \[X\oplus Y\equiv X+Y\mod I^2.\]
For $a\in\CO_K$, $[a].X=F_{aX}\in\CO_K[[X]]$, then
  \[P([a].X)=[a].P(X)\]
and \[a[X]=aX\mod I^2.\]
In particular, $[\pi].X=P$ by unicity.
\begin{thm}
(1) $\oplus$ is a commutative formal group law $\Gamma$, i.e.,
\[X\oplus Y=Y\oplus X,\quad (X\oplus Y)\oplus Z=X\oplus(Y\oplus Z),\quad ([-1].X)\oplus X=0.\]

(2) $a\mapsto [a].X$ is a  ring homomorphism $\CO_K\inj \End\, \Gamma$, i.e.,
  \[[a].(X\oplus Y)=([a].X)\oplus([a].Y),\quad ([a].X)\oplus([b].X)=[a+b].X,\quad [a].([b].X)=[ab].X.\]
\end{thm}
\begin{proof}
Since \[(X\oplus Y)\oplus Z\equiv X+Y+Z\equiv X\oplus (Y\oplus Z)\mod I^2,\]
\[P((X\oplus Y)\oplus Z)=P(X)\oplus P(Y)\oplus P(Z)=P(X\oplus (Y\oplus Z)),\]
we have $(X\oplus Y)\oplus Z=X\oplus(Y\oplus Z)$ by unicity. Similar for other results.
\end{proof}
$(\Gamma,\oplus)$ is a Lubin-Tate formal group attached to $(K,\pi)$.
\begin{prop}
(1) If $P_1,P_2$ as above, then there is a unique $G\in X+\pi^2 \CO_K[[X]]$ such that $G(P_1(X))=P_2(G(X))$.

(2) $G(X\oplus_1 Y)=G(X)\oplus_2 G(Y), G([a]_1. X)=[a]_2 .G(X)$, i.e., $G$ is an isomorphism $(\Gamma_1,\oplus_1)\simto (\Gamma_2,\oplus_2)$.
\end{prop}
\begin{proof}
By unicity.
\end{proof}

\begin{example}
$K=\BQ_p$, $P=(1+X)^p-1$, then
\[X\oplus Y=(1+X)(1+Y)-1,\quad [a].X=(1+X)^a-1,\]
i.e., the multiplicative formal group $\hGm$.
\end{example}
\begin{remark}
A formal group law over $\CO_K$ turns $\fm_{\BC_p}$ into a group.
\end{remark}
\begin{thm}
Let $(\Gamma,\oplus)$ be the Lubin-Tate formal group attached to $(K,\pi)$, define the Tate module
  \[T_\pi(\Gamma)=\{(0,u_1,u_2,\ldots): u_n\in \fm_{\BC_p},[\pi]u_{n+1}=u_n\}.\]

(1) $T_\pi (\Gamma)$ is an $\CO_K$-module of rank $1$.

(2) If $(0,u_1,\ldots)$ is a generator (i.e., $u_1\neq 0$), then $K_n=K(u_n)$ is a totally ramified abelian extension of $K$ with Galois group $(\CO_K/\pi^n)^\times$, where $v_i(u_n)=\frac{1}{(q-1)q^{n-1}}v_p(\pi)$.

(3) Let $K_\infty=\cup K_n$, then $\Gal(K_\infty/K)=\CO_K^\times$. Let $\chi_L:G_K\ra\Gal(K_\infty/K)\ra \CO_K^\times$ be the Lubin-Tate character, then $\sigma(u_n)=[\chi_L(\sigma)]. u_n$.
\end{thm}
\begin{remark}
(1) For $(\BQ_p,p)$, $\Gamma=\hGm$, this becomes the cyclotomic theory.

(2) By local class field theory,
  \[1\ra \CO_K^\times\ra G^{\ab}_K\ra \Gal(\bar \BF_p/\BF_q)\ra 1, \]
thus $K^{\ab}=\cup_{(N,p)=1}K_\infty(\mu_N)$. Lubin-Tate makes LCF completely explained.
If $[K:\BQ]<\infty$, we have a description of $G^\ab_K$ but not of $K^\ab$ (Hilbert's 12th problem).

(3) $T_p(\Gamma)\simeq T_\pi(\Gamma)$, $p=\pi^e a,a\in\CO_K^\times$. If $(u_n)\in T_\pi(\Gamma)$, then $(u_n=[a^{-n}].u_{en})\in T_p(\Gamma)$.
\end{remark}
\begin{proof}
For $a\in\CO_K,(u_n)\in T_\pi(\Gamma), ([a]. u_n)\in T_\pi(\Gamma)$ makes $T_\pi(\Gamma)$ a $\CO_K$-module.
We can assume $[\pi].X=\pi X+X^q$. Then $T_\pi(\Gamma)$ has no $\pi$-torsion. $u\in[\pi].T_\pi(\Gamma)$ iff $u_1=0$, thus $u\mapsto u_1$ injects \[T_\pi(\Gamma)/\pi T_\pi(\Gamma)\inj \Gamma[\pi]=\{x: \pi x+x^q=0\}.\]
Thus $T_\pi(\Gamma)$ has rank $\le 1$ with equality if it is not $0$.

If it is not $0$,  $u$ is a generator iff $u_1\neq 0$, $u_1$ is a solution of $u_1^{q-1}+\pi=0$ and $u_{n+1}$ is a solution of $u_{n+1}^q+\pi u_{n+1}=u_n$, where $X^q+\pi X-u_n$ is Eisenstein. By induction, we get $K_n/K$ is totally ramified and $\pi_n$ is a uniformizer.

\[T_\pi(\Gamma)/\pi^n T_\pi(\Gamma)\simeq \Gamma[\pi^n]\simeq \CO_K/\pi^n.\]
Since $u_n\in \Gamma[\pi^n]-\Gamma[\pi^{n+1}]$, for $\sigma\in G_K$, $\sigma([\pi]-x)=[\pi].\sigma(x)$,
$\sigma(u_n)\in \Gamma[\pi^n]-\Gamma[\pi^{n+1}]$, thus there is $\chi_{L,n}(\sigma)\in(\CO_K/\pi^n)^\times$ such that $\sigma(u_n)=[\chi_{L,n}(\sigma)].u_n$. Hence $\Gal(K_n/K)\simto (\CO_K/\pi^n)^\times$ and $\chi_L=\plim \chi_{L,n}:\Gal(K_\infty/K)\simto \CO_K^\times$.
\end{proof}
$\chi_L:G_K\ra \CO_K^\times$ is  a $1$-dimensional representation of $G_K$ over $K$, then it is a $h$-dimensional representation of $G_K$ over $\BQ_p$. Going to prove that this representation if de Rham, denote $V_\pi(\Gamma)=K\otimes_{\CO_K}T_\pi(\Gamma)$, $\Hom_{G_K}(V_\pi(\Gamma),B_\dR^+)$ is of dimensional $h$. We are going to prove that using ``periods'' of Lubin-Tate formal groups.

Define the logarithm
  \[\partial f(X)=\frac{f(X\oplus Y)-f(X)}{Y}|_{Y=0},\]
then if $t_a^*f(X):=f(X\oplus a)$, $t_a^*\circ \partial=\partial\circ t_a^*$. We have $\partial f(X)=u(X)\frac{\diff f}{\diff X}(X)$
where $u(X)=(\frac{X\oplus Y-X}{Y})_Y\in 1+X\CO_K[[X]]$. Write
  \[\frac{1}{u(X)}=1+a_1X+a_2X^2+\cdots,\]
let
  \[\ell(X)=\int \frac{\diff X}{u(X)}=X+a_1\frac{X^2}{2}+\cdots.\]
$\ell(X)\notin \CO_K[[X]]$ but it converges on $\fm_{\BC_p}$. We have $\ell(X\oplus Y)=\ell(X)+\ell(Y)$. $\ell$ is the logarithm of $(\Gamma,\oplus)$ and
  \[X\oplus Y=\ell^{-1}(\ell(X)+\ell(Y)).\]
\begin{example}
For $\Gamma=\hGm$, $u(X)=1+X$ and $\ell(X)=\log(1+X)$.
\end{example}
We have $\ell([a].X)=a\ell(X)$ if $a\in\CO_K$.
\begin{thm}[Cartier-Harda]
$\ell(X)=\sum_{n\ge 1}\frac{X^{q^n}}{\pi^n}$ is the logarithm of a Lubin-Tate attached to $(K,\pi)$.
\end{thm}
Let $P=X^q+\pi X$, $Q_0=X^{q-1}+\pi$, $Q_{n+1}=Q_n\circ P$.
\begin{prop}
$\ell(X)=X\prod_{n\ge 0}\frac{Q_n}{\pi}$.
\end{prop}
\begin{proof}
$Q_n=\pi+a_{n,1}X+\cdots$, then
\[Q_{n+1}=\pi+a_{n,1}(X^q+\pi X)+\cdots.\]
$v_p(a_{n,q})$ tends to zero. Thus  $\pi^{-1}{Q_n}-1$ tends to zero, and the product converges.

Let $F=X\prod \frac{Q_n}{\pi}$, then $F\circ P=\pi F$ and $\ell\circ P=\pi \ell$, thus
\[(F-\ell)(P)=\pi(F-\ell)\]
and $F-\ell=a_2 X^2+\cdots$, and we have $F=\ell$.
\end{proof}
We have that the zeroes of $\ell$ are exactly $\Gamma[\pi^\infty]$.

\subsection{Periods of Lubin-Tate groups}
Assume $K/\BQ_p$ is Galois, $g\in\Gal(K/\BQ_p)$. There is a unique $0\le i\le f-1$ such that $g(x)=x^{p^i}$ on $k_K$. Then
$\ell_g(X)=g(\ell(X^{p^i}))$ if
\[\begin{split}
  \ell(X)&=X+a_2X^2+\cdots,\\
  \ell_g(X)&=X^{p^i}+g(a_2)X^{2 p^i}+\cdots.
\end{split}\]
\begin{lemma}
(1) $\ell_g(X\oplus Y)-\ell_g(X)-\ell_g(Y)\in \pi^{-N}\CO_K[[X,Y]]$ (quasi-logarithm).
(2) $\ell_g([a].X)-g(a)\ell_g(X)\in \pi^{-N} \CO_K[[X]]$ for $a\in\CO_K$.
\end{lemma}
\begin{proof}
(1) We have $g(X^{p^i}\oplus Y^{p^i})-(X\oplus Y)^{p^i}=\pi R$ for $R\in \CO_K[[X,Y]]$ because $g(x)\equiv x^{p^i}\bmod\pi$ and $x\mapsto x^{p^i}$ is a ring homomorphism.
\[\ell_g(X\oplus Y)=g(\ell((X\oplus Y)^{p^i}))=(g\circ \ell)(g(X^{p^i\oplus}Y^{p^i})-\pi R).\]
Now use the Taylor expansion. Let $F=\ell'\in\CO_K[[X]]$, notice that $g(\ell(X^{p^i}\oplus Y^{p^i}))=\ell_g(X)+\ell_g(Y)$, we have
\[\ell_g(X\oplus Y)-\ell_g(X)-\ell_g(Y)=\sum_{n\ge 1} g(F^{[n-1]}(X^{p^i}\oplus Y^{p^i}))\frac{\pi^n}{n}R\]
where $F^{[k]}:=\frac{1}{k!}F^{(k)}$. Since $(X^a)^{[k]}={a\choose k}X^{a-k}$, $F^{[k]}$ preserves integral coefficients. Thus there  is $N$ such that $\frac{\pi^n}{n}\in\pi^{-N}\CO_K$ and then

(2) is similar to (1).
\end{proof}

\begin{prop}
$u\in T_\pi(\Gamma), \hat u_n\in\wt A^+$ with $\theta(\hat u_n)=u_n$, then $g(\pi)^n\ell_g(\hat u_n)$ has a limit $\int_u \diff \ell_g$ in $B_\dR^+$, which is nonzero for nonzero $u$. Moreover, for $\sigma\in G_K$, $\sigma(\int_u \diff \ell_g)=g(\chi_L(\sigma))\int_u \diff \ell_g=\int_{\sigma(u)} \diff\ell_g$. Thus $\ell_g\in\Hom_{G_K}(T_\pi(\Gamma),B_\dR^+)$ spans a dimension $[K:\BQ_p]$ vector space, which implies that $T_\pi(\Gamma)$ is de Rham.
\end{prop}
\begin{proof}
Let $K_0=W(k_K)[\frac{1}{p}]$. Consider
  \[\begin{split}\theta:\CO_K\otimes_{\CO_{K_0}}\wt A^+&\ra\CO_{\BC_p}\\
  \theta(\sum_{i\ge 0}[x_i]\pi^i)&=\sum x_i^\sharp \pi^i.
  \end{split}\]
Then $\ker \theta$ is generated by $\varpi=[\pi^\flat]-\pi$. Since
  \[\theta([\pi].\hat u_{n+1})=[\pi].\theta(\hat u_{n+1})=[\pi].u_{n+1}=u_n,  \]
we have $[\pi].\hat u_{n+1}=u_n+x\varpi$ for some $x$.
  \[g(\pi)^{n+1}\ell_g(\hat u_{n+1})-g(\pi)^n\ell_g(\hat u_n)=g(\pi)^n(g(\pi)\ell_g(\hat u_{n+1})-\ell_g([\pi].\hat u_{n+1}-x\varpi)).\]
By Lemma,
  \[g(\pi)\ell_g(\hat u_{n+1})-\ell_g([\pi].\hat u_{n+1})\in \pi^{-N}(\CO_K\otimes \wt A^+).\]
Now
  \[\ell_g([\pi].\hat u_{n+1}-x\varpi)-\ell_g([\pi].\hat u_{n+1})\in \sum_{n\ge 1}\frac{\varpi^n}{n} (\CO_K\otimes\wt A^+)\in\pi^{-N(k)}(\CO_K\otimes\wt A^+)+\varpi^{k+1}B_\dR^+\]
bounded mod $\varpi^{k+1}$ for any $k$, thus bounded in $B_\dR^+$. Hence $\ell_g(\hat u_n)$ is bounded and $g(\pi)^n\ell_g(\hat u_n)$ tends to zero.

By this, the limit is independent of the choice of $\hat u_n$. We may take $\wh{\sigma(u_n)}=\sigma(\hat u_n)$ and then
\[\sigma(\int_u \diff \ell_g)=g(\chi_L(\sigma))\int_u \diff \ell_g=\int_{\sigma(u)} \diff\ell_g.\]

Since $[a].\hat u_n=\wh{[a].u_n}+x\varpi$, by Lemma,
  \[\ell_g([a].\hat u_n)-g(a)\ell_g(\hat u_n)\in\pi^{-N}\CO_K\otimes\wt A^+. \]
Then
  \[\ell_g([a].\hat u_n+x\varpi)-\ell_g([a].\hat u_n)\in\sum_{n\ge 1}\frac{\varpi^n}{n})(\CO_K\otimes\wt A^+)\]
is bounded. The rest part is similar.

For $u=(0,u_1,\ldots)\in T_\pi(\Gamma)$ with $u_1\neq 0$, i.e., $u$ is a generator of $T_\pi(\Gamma)$, then
  \[v_p(u_n)=\frac{1}{(q-1)q^{n-1}}v_p(\pi).\]
Since
  \[\ell(X)=\frac{\stackrel{n-1}{\overbrace{P\circ P\circ\cdots \circ P}}}{\pi^{n-1}}\frac{Q_n}{\pi}\prod_{k\ge n}\frac{Q\circ P^k}{\pi}.\]
Since the Eisenstein polynomial $Q_n$ is the minimal polynomial of $u_n$ over $K$, $Q_n(\hat u_n)\in\ker \theta$ is a generator. Thus
  \[v_p(\theta(\frac{Q_n(\hat u_n)}{\varpi}))=0.\]
Since
  \[\frac{\theta(Q\circ P^k(\hat u_n))}{\pi}=\frac{Q\circ P^k(u_n)}{\pi}=Q(0)/\pi=1.\]
  \[v_p(P\circ \cdots\circ P(u_n))=v_p([pi^{n-1}].u_n)=v_p(u_1)=\frac{v_p(\pi)}{q-1}.\]
Since the valuation of $\theta(\pi^n\frac{\ell(\hat u_n)}{\varpi})$ is $v_p(\pi)+\frac{1}{p-1}v_p(\pi)$ is independent of $n$,
  \[\theta(\pi^n\frac{\ell(\hat u_n)}{\varpi})\ra \theta(\frac{\int_u \diff \ell}{\varpi})\]
is nonzero.
\end{proof}

Let $K/\BQ_p$ be a finite Galois extension, $(\Gamma,\oplus)$ be a dimension $d$ commutative formal group, that is, for $X=(X_1,\ldots,X_d), Y=(Y_1,\ldots,Y_d)$,
  \[X\oplus Y=((X\oplus Y)_1,\ldots,(X\oplus Y)_d)\]
with $(X\oplus Y)_d\in\CO_K[[X,Y]]$ and $(X+Y)_i\equiv X_i+Y_i\bmod \deg 2$, such that
\[X\oplus Y=Y\oplus X,\]
\[(X\oplus Y)\oplus Z=X\oplus (Y\oplus Z).\]

We can get a true group on $(\fm_{\BC_p})^d=B_d(0,1^-)$. We have a rank $k$ Galois $\BZ_p$-module $T_p(\Gamma)$.


Let
  \[\RH^1_\dR(\Gamma)=\frac{\{\omega\in(\Omega^1_{\CO_K[[X]]})^{\diff=0}:
  F_\omega(X\oplus Y)-F_\omega(X)-F_\omega(Y)\in K\otimes\CO_K[[X]]\text{ for }\diff F_\omega=\omega\}}{\{\diff F:F\in K\otimes \CO_K[[X]]\}}.\]
We can write $\omega=f_1\diff x_1+\cdots+f_d \diff x_d$ for $f_i\in\CO_K[[X]]$.
\begin{thm}
(1) $\dim_K\RH^1_\dR(\Gamma)=k=\dim_{\BZ_p}T_p(\Gamma)$.

For $\omega$ quasi-log, $(u_n)\in T_p(\Gamma)$, $\hat u_n\in(\wt A^+)^d$, $\theta(\hat u_n)=u_n$, the limit of $p^n F_\omega(\hat u_n)$ exists and does not depend on $\hat u_n$, which is called the period $\int_u \omega\in B_\dR^+$ of $\omega$. It's zero for $\omega=\diff F$ for some $F\in K\otimes\CO_K[[X]]$.

(2) \[\Function{\RH^1_\dR(\Gamma)\times T_p(\Gamma)}{B_\dR^+}{(\omega,u)}{\int_u\omega}\]
is linear, commutes with $G_K$-action. It respects filtrations if $\omega \in \Omega^1_\inv(\Gamma)$, then $\int_u\omega\in t B^+_\dR$.
  \[\RH^1_\dR(\Gamma)\inj \Hom_{\CO_K}(T_p(\Gamma),B^+_\dR)\]
implies $T_p(\Gamma)$ is de Rham.
\end{thm}

\subsection{$p$-adic integration}
Assume $[K:\BQ_p]<+\infty$, $X/K$ a smooth projective curve with Jacobian $J$. Fix $\iota:X\ra J$. For $\omega\in\Omega^1_{K(X)}$, we want to define $F_\omega=\int \omega$, which satisfies
\begin{enumerate}
\item $F_\omega$ locally analytic outside the poles of $\omega$;
\item $\diff F_\omega=\omega$.
\end{enumerate}

In the complex case, $F_\omega$ will be multivalued. But in the $p$-adic world, $F_\omega$ can be defined around each point, but no analytic continuation  because balls are  disjoint. There will be two many $F_\omega$ because of the locally constant functions. On abelian varieties, the group structure will help figure out the $F_\omega$ we want. So, for general varieties, we will define the p-adic integral theory using their Albanese varieties.

For $\log=\int \frac{\diff x}{x}$, choices made smaller by requiring
  \[\log xy=\log x+\log y, \]
and
\[\diff\log=\id: \BG_a\rightarrow \BG_a.\]
If furthermore fix $\log p=\CL$, we will get a unique $\log$ denote by $\log_\CL$.

Let $Z=X$ or $J$. There is an exact sequence

\[\xymatrix{
0\ar[r]&\RH^0(Z,\Omega^1)\ar[r]&\DSK(Z)\oplus (K\otimes \DTK(Z))\ar[r] &(\Omega^1_{K(Z)})^{\diff=0}\ar[r]&0
}\]

We want $\int \diff f=f$ and $\int \frac{\diff f}{f}=\log_\CL f$ up to global constants.

Recall that there is a bijection of sets:
\[\iota^*:(\Omega^1_{K(J)})^{\diff=0}/\{\text{exact}\}\simto\Omega^1_{K(X)}/\{\text{exact}\}\]
and there are three maps $m,\pr_1,\pr_2$ from $J\times J$ to $J$.
\begin{thm}[Theorem of square]
For $\omega\in(\Omega^1_{K(J)})^{\diff=0}$,
  \[m^*\omega-\pr^*_1\omega-\pr^*_2\omega\]
is exact on $J\times J$, and can be written as $\diff F_\omega^{(2)}(x,y)$, where
  \[F_\omega^{(2)}(x,y)=F_{0}(x,y)+\sum \lambda_i\log_\CL F_{i}(x,y)\]
up to constant with $F_{0}(x,y)\in K(J\times J)$ and $F_{i}(x,y)\in {K(J\times J)}^*$.
\end{thm}

\begin{thm}[Main theorem of integration]
If $\omega \in (\Omega^1_{K(J)})^{\diff=0}$, then there exists a unique $F_\omega$ locally analytic on $J(\BC_p)$ with $\diff F_\omega=\omega$ and
  \[F_\omega(X\oplus Y)-F_\omega(X)-F_\omega(Y)=F_\omega^{(2)}(X,Y).\]
\end{thm}
Main step of the proof:
\begin{enumerate}
\item[(A)] $J(\BC_p)$ contains a basis $\{U_i\}$ of neighborhood of $0$ consists of open subgroups. Furthermore, $J(\BC_p)/U_i$ is a torsion group for any $i$ (proved by formal groups).
\item[(B)] Formal integral $\omega$ to get an analytic function $F_\omega$ on a small enough open subgroup $U$ of $J$. Then using the function $F_\omega^{(2)}$ which is constructed by square theorem to continuous $F_\omega$ to $J$ and satisfy the relation in the theorem.


By theorem of square, $\exists$ $F_\omega^{(2)}(x,y)=F_{0}(x,y)+\sum \lambda_i\log_\CL F_{i}(x,y)$ such that $dF_\omega^{(2)}=m^*\omega-\pr^*_1\omega-\pr^*_2\omega$.
\end{enumerate}

\begin{proof}
(1)By the Theorem below.

(2)For a closed form $\omega\in(\Omega^1_{K(J)})^{\diff=0}$, let $F_\omega^{(2)}$ be the function on $J\times J$ as in the Theorem of square.
By formally integrality, we get $F_\omega$  analytic globally on some neighborhood $U$ of zero,
  \[F_\omega=F_0+\sum \lambda_i \log_\CL F_i.\]
We can take $U$ to be a subgroup.

If $\omega\in\RH^0(J,\Omega^1)$, we  can take $F_\omega^{(2)}=0$. We want $F_\omega([a].x)=aF_\omega(x)$. For $x\in J(\BC_p)$, there is $m$ such that $[m].x\in U$, we take $F_\omega(x)=\frac{1}{m}F_\omega([m].x)$. Since $F_\omega$ is analytic on $U$,
  \[F_\omega(x\oplus y)-F_\omega(x)-F_\omega(y)\]
is analytic on $U\times U$ and $\diff =0$, thus it is zero on $U\times U$ and we get the formula.

In general case, let $f_2(x)=F_\omega^{(2)}(x,y)=F_\omega([2].x)-2F_\omega(x)$ on $U$. Let
  \[f_n(x)=f_{n-1}(x)+F_\omega^{(2)}([n-1].x,x)=F_\omega([n].x)-n F_\omega(x)\]
on $U$, then
  \[f_{n,m}(x)=f_n([m].x)+n f_m(x)=F_\omega([nm].x)-nm F_\omega(x).\]
Define
  \[F_\omega(x)=\frac{1}{n}(F_\omega([n].x)-f_n(x))\]
with $n$ such that $[n].x\in U$, then it does not depend on $n$. This finishes the proof.
\end{proof}

\begin{remark}
For $\omega\in\RH^0(J,\Omega^1)$, $m^*\omega=\pr_1^*\omega+\pr_2^*\omega$, we can take $F_\omega^{(2)}=0$ and
  \[F_\omega(X\oplus Y)=F_\omega(X)+F_\omega(Y).\]
It's called the logarithm of $J$.
\end{remark}

\begin{thm}
(1) $J(\BC_p)$ contains a basis of neighborhood of $0$ of open subgroups.

(2) If $U$ is one of these open subgroups, $J(\BC_p)/U$ is a torsion group.
\end{thm}
\begin{proof}
Let $x_1,\ldots,x_g\in K(J)$, $\diff x_i-\omega_i$ vanishes at $0$, $z\mapsto (x_1(z),\ldots,x_g(z))$ is an analytic isomorphism between some neighborhood of $0$ and $B_d(0,\delta)^-=\{x\in\BC^d\mid v_p(x_i)>\delta\}$.
Then
\[x_i(z_1\oplus z_2)=x_i(z_1)+x_i(z_2)+F_i(x(z_1),x(z_2))\]
for $F_i\in (x(z_1),x(z_2))^2 K[[x(z_1),x(z_2)]]$ converges in $B_{2d}(0,\delta^-)$ for $\delta^->\delta$.

Let $M=\inf_i v_p(F_i(x,y))$, $(x,y)\in B_{2d}(0,\delta^-)$, then
  \[v_p(p^k x,p^k y)\ge 2k+M\]
if $(x,y)\in B_{2d}(0,\delta^-)$. If $k+M\ge \delta^-$, $v_p(F_i(p^k x,p^ky))\ge k+\delta^-$, thus $B_{2d}(0,k+\delta')$ is stable by $\oplus$, and neighborhood is a group. For any $k$ big enough, the inverse image of $B_{2d}(0,k+\delta^-)$ is an open subgroup of $J(\BC_p)$.

Since $\ov \BQ_p$ is dense in $\BC_p$,
  \[J(\ov\BQ_p)/(U\cap J(\ov\BQ_p))\simeq J(\BC_p)/U,\]
where $J(\ov\BQ_p)=\bigcup\limits_{[L:K]<\infty} J(L)$. Since $J(L)$ is a compact group, the image of $J(L)$ in $J(\BC_p)/U$ is finite, thus it is torsion and then so $J(\BC_p)/U$ is.

The compactness of $J\subset \BP^d$ follows from that $\BP^d(L)$ is compact since it is a union of some
  \[\bigcup_{i=0}^d \CO_L\times\cdots\times\CO_L\times 1\times\CO_L\times \cdots\times\CO_L,\]
and $\CO_L$ is compact because $[L:\BQ_p]<\infty$.
\end{proof}
\begin{remark}
If $X$ has a good model over $\CO_K$, then $J$ also has a good model $\fJ$. Moreover,
  \[0\ra U\ra \fJ(\CO_{\BC_p})\ra \fJ(\ov\BF_p)\ra 0,\]
where $U$ is analytically the unit open ball $B_g(0,0^-)$. $\oplus$ on $J$ gives an addition law on $B_g(0,0^-)$ and $(x\oplus y)_i\in\CO_K[[x,y]]$ gives a formal group law defined over $\CO_K$.
\end{remark}


\subsection{$p$-adic periods of abelian integrals}
Recall $\RH^1_\dR=\frac{\DSK(Z)}{\{\diff f\}}$ and the pairing
\[\Function{\RH^1_\dR(Z)\times H_1(Z(\BC),\BZ)}{\BC}{(\omega,u)}{\int_u\omega.}\]
For $\omega\in \DSK(J)$, $U\subset J$ affine open on which $\omega$ is holomorphic. Write $U=\Spec(K[x_1,\ldots,x_n]/I)\inj \BA^n$. Say $A\subset U(B_\dR^+)$ is bounded if its projection on each $\BA^1$ is bounded in $B_\dR^+$, i.e,
for any $k$, $\exists N(k)$ such that $x_i(A)\subset p^{-N(k)}(\wt A^+\otimes\CO_K)+(\ker\theta)^{k+1}$.

Define the Tate module
\[T_p(J):=\{(0,u_1,\ldots):u_n\in J(\BC_p),[p].u_{n+1}=u_n\}.\]
\begin{thm}[$p$-adic periods]
(1) We can find bounded sequences $(a_n),(b_n)$ in $U(B_\dR^+)$ with $\theta(b_n)\ominus\theta(a_n)=u_n$.

(2) $p^n(F_\omega(b_n)-F_\omega(a_n))$ has a limit $\int_u\omega\in B_\dR^+$, which depends only on $u$ and the image of $\omega$ in $\RH^1_\dR(J)$. Thus we have a pairing
  \[\Function{\RH^1_\dR(J)\times T_p(J)}{B^+_\dR}{(\omega,u)}{\int_u\omega.}\]
It is $G_K$-equivariant,
  \[\int_{\sigma(u)}\omega=\sigma(\int_u\omega),\]
respects filtration. For $\omega\in\RH^0(J,\Omega^1)$, $\int_u \omega\in tB_\dR^+$.

(3) \[\RH^1_\dR(J)\lra \Hom_{G_K}(T_p(J),B^+_\dR)\]
is injective and therefore $\BQ_p\otimes T_p(J)$ is de Rham.
\end{thm}
\begin{proof}
The non-degenerate is a consequence of Riemann relation.
\end{proof}

Idea behind the construction of $p$-adic periods $\int_u \omega=\lim p^n F_\omega(\hat u_n)$:
We say a function natural if it's bounded outside their poles, that is, $f$ holomorphic on $U=\Spec(K[x_1,\dots,x_n]/T)$, $f$ is bounded on any bounded set in $U$.
For example, $\frac{1}{1+x}$ is bounded on $v_p(x)\ge 0$ and $v_p(1+x)\ge 0$, but $\log(1+x)$ is not bounded  on $v_p(x)>0$.

If $\omega\in \DSK$, $F_\omega([p].x)-pF_\omega(x)$ is natural.
\[p^{n+1}F_\omega(\hat u_{n+1})-p^{n}F_\omega(\hat u_{n})=p^n(pF_\omega(\hat u_{n+1})-F_\omega([p].\hat u_{n+1})+F_\omega([p].\hat u_{n+1})-F_\omega(\hat u_n)).\]
Use Taylor expansion, we get the naturality.

More conception construction.

(1) Recall the universal extension
\[0\ra \RH^0(J,\Omega^1)\ra \wt J\sto{\pi}J\ra 0.\]
For $\omega\in \DSK(J)$, there exists a unique $\eta(\omega)\in \RH^0(\wt J,\Omega^1)$ invariant by translation, such that $\pi^* \omega-\eta(\omega)=\diff f$ for some $f\in K(\wt J)$. We can define $F_{\eta(\omega)}$ by $\frac{1}{n}F_{\eta(\omega)}([n].x)$, then we get a formula for $F_\omega$.

(2) Let
  \[\hat J(\BC_p)=\{u=(u_0,u_1,\dots,u_n,\dots) : u_n\in J(\BC_p), [p].u_{n+1}=u_n\},\]
then
  \[0\ra T_pJ\ra \hat J(\BC_p)\xrightarrow{u\mapsto u_0} J(\BC_p) \ra 0.\]
This is an analogue
\[0\ra \RH_1(J(\BC),\BZ)\ra \BC^g\ra J(\BC)\ra 0.\]

$u\in \hat J(\BC_p)$, $\hat u_n\in \wt J(B_\dR^+)$ bounded with $\pi(\theta(\hat u_n))=u_n$. then $[p^n].\hat u_n$ converges to $\iota_\dR(u)$ in $\wt J(B^+_\dR)$. For $u\in T_pJ$, $\int_u \omega=F_{\eta(\omega)} (\iota_\dR(u))$.

\subsection{$p$-adic Riemann relations}
Let $\omega_1,\ldots,\omega_g$ be a basis of $\RH^0(J,\Omega^1)$, $\pi:\BC^g\ra J(\BC)$ the projection. Then
  \[\diff f=\sum_{i=1}^g \partial_i f\omega_i,\]
where $\partial_i$ are translate invariant differential operators.
For the theta function $\theta$ on $\BC^g$, $\wt\eta_i=\diff\,(\frac{\partial_i\theta}{\theta})$ comes from a differential form $\eta_i$ on $J$, i.e., $\pi^*\eta_i=\wt\eta_i$ for $\eta_i\in \DSK(J)$.
Then $\omega_1,\ldots,\omega_g,\eta_1,\ldots,\eta_g$ is a basis of $\RH^1_\dR(J)$. Moreover
\[\sum_{i=1}^g \int_u \eta_i\int_v \omega_i -\int_v\eta_i\int_u \omega_i=2\pi i (u\# v).\]
The theorem of the cube says
  \[\frac{\theta(z_1+z_2+z_3)\theta(z_1)\theta(z_2)\theta(z_3)}{\theta(z_1+z_2)\theta(z_2+z_3)\theta(z_3+z_1)}=\pi^* f_x(x_1,x_2,x_3),\quad f_x\in\BC(J\times J\times J)^\times.\]

In $p$-adic case, we can define $\log_\CL\theta$ with $\diff \log_\CL\theta=\sum_{i=1}^g F_{\eta_i}\omega_i$ by Green function.
\begin{thm}
There exits a Green function $G$ unique up to a polynomial of degree $2$ in the logarithm of $J$, such that
  \[\sum_{\emptyset\neq S\subseteq\{1,2,3\}}(-1)^{\#S}G(\bigoplus_{i\in S} x_i)=\log_\CL f_x(x_1,x_2,x_3).\]
\end{thm}

The Weil pairing
  \[\pair{-,-}_\Weil: T_p(J)\times T_p(J)\ra T_p(\mu_{p^\infty})=\BZ_p t,\]
where $T_p(J)=\BZ_p\otimes\RH_1(J(\BC),\BZ)$, $\pair{u,v}_\Weil=(u\# v)t$. It is a big theorem that Weil pairing is non-degenerate.
\begin{thm}
\[\sum_{i=1}^d\int_u \eta_i\int_v \omega_i-\int_v \eta_i\int_v\omega_i=\pair{u,v}_\Weil.\]
\end{thm}
Since $\pair{-,-}_\Weil$ is non-degenerate, $\RH^1_\dR(J)\inj \Hom_{G_K}(T_p(J),B^+_\dR)$.

\subsection{One example of application}
Let $K$ be a number field, and $X/K$ be a smooth proper curve, then $J(K)$ is of the type finite group$\times \BZ^n$. Assume that $n\le g-1$, then $X(K)$ is finite (special case of Mordell, Chabauty's method). Let $P-1,\ldots,P-n\in J(K)$ such that $J(K)/\pair{P_1,\ldots,P_n}$ is torsion, then
Since
  \[\dim \RH^0(J,\Omega^1)=g>n,\]
there is a nonzero $\omega\in \RH^0(J,\Omega^1)$ such that $F_\omega(P_1)=\dots=F_\omega(P_n)=0$, $F_\omega(0)=0$, thus $F_\omega(P)=0$ for any $P\in J(K)$.
For $P_0\in X(K)$, $\iota_{P_0}: X\ra J$, $\iota(X(K))\subset J(K)$. For $f=F_\omega\circ \iota_{P_0}$ locally analytic function on $X$, $f(P)=0$ for any $P\in X(K)$. Since $X(K_p)\supset X(K)$ is compact, there exists finite set of $U_i$ on which $f$ is analytic and $\cup U_i\supset X(K_p)$, $f$ has a finite number of zeroes on each $U_i$.

\begin{conj}[Caporaso-Harris-Mazur]
For $g\ge 2$, there exists a constant $N(g,K)$ such that for any $X/K$ of genus $g$, $|X(K)|\le N(g,K)$.
\end{conj}
Stoll and Rabinoff proved the case $n\le g-2$ under some technical assumptions.






















%\appendix
%\section{Intersection theory}
%$(Poincar\acute{e}\ duality)$ Let $M$ be a compact $n$-dimensional orientable real manifold with fundamental class $[M]$, then
%$$D: H^k(M,\BR)\to H_{n-k}(M,\BR)$$ is isomorphism for all $k$.
%Let $\eta$ be its inverse.
%
%\begin{defn}
%(Intersection number)
%$$\_\#\_: H_k(M,\BZ)\times H_{n-k}(X,\BZ)\to \BZ,$$$$(x,y)\mapsto<\eta_x\cup\eta_y,[M]>.$$
%\end{defn}
%
%Let $X$ be a compact orientable surface, fix an orientable, then there exists a basis $(a_1,...a_g,b_1,...b_g)$ of $H_1(X,\BZ)$ such that \\
%$$a_i\#b_j=\delta_{ij}=-b_j\#a_i,\ a_i\#a_j=0=b_i\#b_j.$$
%
%\cite{1}Let $X$ be a $n$-dimensional compact smooth manifold\\
%$(1)$ $(Poincar\acute{e}\ duality)$ The following pair is perfect:
%$$\int_X\_\wedge\_:H_{dR}^{n-k}(X)\otimes H_{dR}^k(X)\to\BR,\ (\eta,\xi)\mapsto\int_X\eta\wedge\xi.$$\\
%$(2)$ The following natural map is an isomorphism.
%$$H_k(X,\BR)\to H_{dR}^{k}(X)^*,\ u\mapsto\int_u\cdot\ .$$
%$(3)$ Combining $(1)$ with $(2)$, exists is a isomorphism
%$\eta:H_k(X,\BR)\to H_{dR}^{n-k}(X)$ such that $$\forall\ \omega\in H_{dR}^{k} (X),\ \int_u\omega=\int_X\omega\wedge\eta_u.$$\\
% $\eta$ is called the $Poincar\acute{e}\ dual$ map. \\
%$(4)$ For $u\in H_k(X,\BR)$ and $v\in H_{n-k}(X,\BR)$, \\
%$$u\#v=\int_X\eta_u\wedge\eta_v.$$\\
%$(5)$ Let $Z$ be a closed orientable submanifold of dimension $k$, then the $Poincar\acute{e}\ dual$ of Z can be shrunk into any given tubular neighbourhood of $X$.
%
%\begin{remark}The Riemann's period relation is a natural corollary from above facts.
%\end{remark}
%
%\section{Using the universal extension of elliptic curve to define Eisenstein series of weight 1}
%\label{app:eisen}
%\footnote{Edited by Wei He (何伟).}
%Let
%\begin{itemize}
%\item $f$ be a $N$-period odd function from $\BZ$ to $\BZ$;
%\item $E$ be an elliptic curve;
%\item $\omega$ be a holomorphic differential of $E$;
%\item $P$ be a torsion point of $E$ with order exactly $N$.
%\end{itemize}
%Note that $(E,\omega,P)\simeq(\Lambda, \alpha_P).$
%In \cite{2}, the Eisenstein series of weight one is defined by
%$$G_{1,f}(E,\omega,P)=-\frac{1}{2}\sum_{d\ mod N,d\neq 0}\wh{f}(d)A_1(E,\omega,dP),$$
%where $A_1=A_1(\Lambda,\alpha_P)=\zeta(\alpha_P,\Lambda)-\frac{\eta(N\alpha_P,\Lambda)}{N}.$
%
%The $q$-expansion of this Eisenstein series at $(2\pi i(\BZ\oplus\tau\BZ),\frac{2\pi i}{N})$ is
%  \[\frac{1}{2}L(0,f)+\sum_{n\geq 1}q^n\sum_{d|n}f(d).\]
%In fact, $A_1$ can be defined by the universal extension $\wt E=\BC\times\BC/\{(\lambda,\eta(\lambda))\}$ of $E$.
%
%For any torsion point $\alpha_P$ in E, $\alpha_P^{\mathrm{can}}=(\alpha_P,\frac{\eta(N\alpha_P)}{N})$ is the unique torsion point in $\wt E$ lies over $\alpha_P$. And $\varphi_0:z\mapsto [z,\zeta(z,\Lambda)]$ is  a cross section  of the projection $\wt E\sto{\pi} E$ over $E^{\mathrm{aff}}$. Then it follows that
%  \[A_1(E,\omega,P)=A_1(\Lambda,dz,\alpha_P)=\frac{\phi^{-1}(\varphi_0(\alpha_P)-\alpha_P^{\mathrm{can}})}{\omega}.\]
%
%
%
%\begin{thebibliography}{999}
%\bibitem[BT]{1}
%Bott R, Tu L W, \emph{Differential Forms in Algebraic Topology}. Springer, 1982,44--69.
%\bibitem[K]{2}
%Katz N M, \emph{The Eisenstein measure and $p$-adic interpolation}. American Journal of Mathematics, 1977, 99(2):238--311.
%\end{thebibliography}

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